MSRI Summer School: Automorphic forms and theLanglands program

Ed Belk∗(Lectures ≡ 1 mod 4),Jesse Han (Lectures ≡ 2 mod 4),

Sheela Devadas (Lectures ≡ 3 mod 4),Debanjana Kundu (Lectures ≡ 0 mod 4).

July 24 - August 4, 2017

∗Please direct any complaints to Ed: belked [at] math [dot] ubc [dot] ca.

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1 Lecture One

We begin with a brief overview of some of the topics to be covered in this summerschool.

Let N ∈ Z≥1 be an integer and χ :(Z/NZ

)∗ → C× = GL1(C) a Dirichlet charac-ter modulo N , i.e., a group homomorphism. Attached to χ is a Galois representationρχ of the absolute Galois group

ρχ : Gal(Q/Q)→ GL1(C),

arising from the universal property of quotients; specifically, we take the unique liftχ ∈ Gal(Q/Q) which equals the composition of χ with the surjection Gal(Q/Q) →Gal(Q(ζN)/Q

)=(Z/NZ

)∗, where ζN is a primitive Nth root of unity, the isomor-

phism given byn ∈

(Z/NZ

)× 7−→ (ζN 7→ ζnN

).

In our notation, = will denote a canonical isomorphism.In particular, suppose f is a cuspidal modular form for GL2 which is a simulta-

neous eigenvector for the Hecke operators Tp; we will write

Tpf = λpf, λp ∈ C.

It happens that the subfield Ef of C generated by {λp : p prime} is a number field.If ` ∈ Z is prime and λ|` is a prime of Ef , then (following a suggestion of Serre)

Deligne was able to prove a two-dimensional analogue of the above; namely, theconstruction of the map

ρf : Gal(Q/Q

)→ GL2

(Ef,λ

),

where Ef,λ is the localization of Ef at λ.The representation ρf is “attached to f” in some way. More precisely: if f

has level N ≥ 1, weight k ≥ 1, and (Dirichlet) character χ, then ρf is unramifiedaway from N`. Furthermore, if p is prime, p - N`, then ρf (Frobp) has characteristicpolynomial

X2 − λpX + pk−1χ(p).

The Chebotarev density theorem implies that there exists at most one semisimpleρf with this property.

For k ≥ 2, Deligne constructed ρf using etale cohomology with non-trivial coef-ficients; for k = 1, Deligne and Serre used another method.

Certain questions arise from Deligne’s construction. For instance, if p|N`, whatdoes ρf look like locally at p? We break into two cases:

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1. Suppose p|N, p 6= `. Then the answer comes from the so-called local Lang-lands correspondence. This will be explained (hopefully) later this week.

2. Suppose p = `; then we will use the p-adic local Langlands correspondence.

We consider now an easier variant on the same theme. Instead of asking for therepresentation ρf itself, we might ask for its reduction modulo `, i.e. the map

ρf : Gal(Q/Q

)→ GL2(F`),

recalling that F` is the residue field of Ef,λ at λ. It happens that ρf lies in the`-torsion of an appropriate abelian variety.

We might reasonably ask the question: are ρχ and ρf special cases of a moregeneral phenomenon?

Theorem 1.1. (Harris, Lan, Taylor, Thorne, 2013; Scholze, 2017?)Let E be a totally real or CM number field, π a cuspidal, automorphicrepresentation of GLn(AE). Assume that π∞ is “cohomological” (to bethought of for now as a strong “algebraicity” assumption); then thereexists some representation

ρπ : Gal(E/E)→ GLn(Q`)

attached to π in some canonical way.

This theorem is analogous to giving the characteristic polynomial of ρf (Frobp);more details on all these objects will follow in the coming lectures.

The constructions we have seen so far follow a general theme: namely, to associateto a given algebraic or analytic object (i.e. χ, f, π) a representation of a Galois group(i.e. ρχ, ρf , ρπ). We might ask: is it possible to classify the image? That is, given arepresentation ρ of some Galois group in GLn, does there exist some representationρf (say) arising from the algebraic/analytic object, which is isomorphic to ρ?

We will begin by considering the one-dimensional case. Say K/Q is a finite Galoisextension with one-dimensional representation

ρ : Gal(K/Q)→ GL1(C).

When is this ρ isomorphic to some ρχ, for χ a Dirichlet character?By the universal property of quotients, we may assume that ρ is injective; oth-

erwise we replace K by an appropriate subfield. Because the image of ρ lies in

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an abelian group, it follows that Gal(K/Q) is itself abelian – we say that K is anabelian extension of Q. If

χ : Gal(Q/Q) = Gal(Q(ζN)/Q

)→ C×

is the character above and ρχ as before, then ρχ gives rise to an injection

Gal(L/Q) ↪→ C×,

for some subfield L of Q(ζN).Our question now becomes: If K is a number field which is an abelian Galois

extension of Q, does there exist a positive integer N such that K ⊆ Q(ζN)?

Theorem 1.2. (Kronecker-Weber theorem; “explicit case of global classfield theory”): The answer to the above question is “yes.”

Thus: for all continuous ρ : Gal(Q/Q

)→ GL1(C), there exists a character

χ :(Z/NZ

)× → C× such that ρ ∼= ρχ.Moving to the two-dimensional case: suppose f is a cuspidal modular eigenform

as before, then the associated representation ρf : Gal(Q/Q) → GL2(Q`) has thefollowing properties:

1. ρf is absolutely irreducible.

2. ρf is odd, i.e., det(z 7→ z) = −1.

3. ρf is unramified away from a finite set of primes, and is potentially semi-stable(a p-adic Hodge-theoretic condition) at `.

The third property is sometimes abbreviated by saying ρf is geometric, but weprobably won’t use this terminology.

In the early 1990s, Fontaine and Mazur asked the converse question: If ρ :Gal(Q/Q) → GL2(Ql) satisfies the conditions 1., 2., and 3., then is ρ ∼= ρf forsome f?

The answer to this question is now basically known by the work of Kisin andEmerton, but is still being consolidated and published; see “The Fontaine-Mazurconjecture for GL2” and “Local-global compatibility in the p-adic Langlands programfor GL2,” both of which written circa 2012.

In the general case, we ask: If ρ : Gal(E/E)→ GLn(Ql) satisfies certain assump-tions, is it the case that ρ ∼= ρπ as in theorem 1.2?

This is proven by Barnet-Lamb, Gee, Geraghty, and Taylor for many cases, butnot generally.

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Starting with the next talk, we will begin to include far more details and compu-tations.

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2 Lecture Two

The aim of this course is to give us a feeling for the statements of the theorems thatwe just saw.

We’ll see precise definitions and examples, and learn to work with these objects.And maybe we’ll work with groups other than GLn.

2.1 Part 1: the local Langlands Correspondence

(The first lecture was global.)The local Langlands correspondence for GLn /K is, vaguely speaking, a canonical

bijection between

1. Certain typically infinite-dimensional irreducible C-representations of GLn(K),and

2. certain n-dimensional complex representations of a group related to Gal(K/K).

If n = 1, this is the so-called “local class field theory”.

Remark. Local class field theory has sort of a funny history. Most of the results inglobal class field theory were proved in the early 20th century, before p-adic numbershad really been discovered. Various local results were deduced from the global results,but then people realized that it was more natural to prove the local statements firstand then deduce the global statements.

When n > 1, this is the local Langlands conjectures for GLn /K, which are a2000 theorem of Harris and Taylor (proofs are global).

Now we’ll start being rigorous.

2.1.1 Infinite Galois groups

Let’s remind ourselves of what happens in the finite case. Let K be any field, andsay L/K is a finite extension. We say that L/K is Galois if L/K is normal andseparable.

Then Gal(L/K) is the group of field automorphisms of L fixing K pointwise,with #Gal(L/K) = dimK(L) (i.e. the dimension of L as a K-vector space).

Furthermore, there is an inclusion-reversing correspondence (the Galois corre-spondence) between subgroups H ⊆ Gal(L/K) and intermediate extensions M,K ⊆M ⊆ L.

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Alright, now say that K is a field and L/K is an algebraic extension of possiblyinfinite degree. We’re going to convince ourselves that

Gal(L/K) ' lim←−

Gal(M/K),

where M runs over the finite Galois extensions of K in L.We say that L/K is Galois if it’s normal and separable. Set Gal(L/K) to be the

field automorphisms ϕ : L→ L such that φ fixes K pointwise.The key observation about the (possibly infinite) Galois group is that if ϕ is in

Gal(L/K), then ϕ is determined by its values on the elements of L, each of whichlies in a finite extension of K.

That is, if λ ∈ L, then there exists M,L ⊇ M ⊇ K such that M/K is finite andGalois and λ ∈M .

In particular, there is a canonical quotient map Gal(L/K) → Gal(M/K) andϕ(λ) is determined by the image of ϕ in Gal(M/K).

In particular, ϕ is determined by ϕ|M for all M : L ⊇M ⊇ K, where M is finiteand Galois over K.

Therefore, this induces an embedding

Gal(L/K) ↪→∏

L⊇M⊇KM/K finite, Galois

Gal(M/K).

The product topology induces a nontrivial subspace topology on Gal(L/K) (theKrull topology.) (This is really just the topology of pointwise convergence.)

Exercise 1. Check thatGal(L/K) is a closed subspace of this product.

Of course, no recap of Galois theory is complete without mentioning the fun-damental theorem of Galois theory: there is an inclusion-reversing bijective corre-spondence between the closed subgroups of Gal(L/K) and the intermediate fieldsL ⊇M ⊇ K.

Example 1. Here are some examples.

1. Let K = Q and let L =⋃n≥1 Q(ζpn), where ζpn is e2πi/pn , p prime. Set

Ln = Q(ζpn). We know that Gal(Ln/Q) is (Z/pnZ)×, and so

Gal(L/Q) ↪→∏n≥1

(Z/pnZ)× ,

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sending ϕ 7→ (ϕn)n≥1. There is an infinite tower

Q ⊆ L1 ⊆ L2 ⊆ L3 ⊆ · · · ⊆ Ln ⊆ . . . ,

with⋃Li = L and each ϕn : Ln → Ln.

Hence, if we know ϕn, we know what ϕm is for all m ≤ n. More precisely, ifϕ� (ϕn), and ϕn ∈ (Z/pnZ)×, then ϕm = ϕn modulo pm.

In particular, Gal(L/Q) = lim←−

(Z/pnZ)× = Z×p , equipped with the subspace

topology from Zpdf= lim←−

Z/pnZ.

2. Let’s do an example closer to what we’ll be interested in: local fields. Well,before we do local fields, let’s do a finite field.

Let K be finite, and let L = K. Say that #K = q. Then L is the union of theunique qnth-order extensions of K.

Remark. Note that Fq ⊆ Fq2 , but Fq2 6⊆ Fq3 , simply by dimension + countingelements of a finite-dimensional vector space over a finite field. How it worksis that this inclusion only happens if the smaller exponent divides the larger.

Let’s remind ourselves how Galois theory over finite fields works. Write Lndf=

Fqn . Ln/K is Galois over n and one checks that Gal(Ln/K) = Z/nZ where thegenerator 1 on the right hand side goes to Frobq, which is a completely explicitmember of the Galois group of Ln/K.

So, we can send g ∈ Gal(K/L) to (gn) ∈∏

Z/nZ.

(gn) is in the image of Gal(K/K), if and only if gn modulo m is gm for all mdividing n.

Therefore, Gal(K/K) is lim←−

Z/nZ = Z =∏

p Zp, the limit taken over the

inverse system of positive integers, ordered by divisibility.

3. Now, local fields. Let’s stick to the case where K/Qp is finite. Choose analgebraic closure K over K. We want to understand Gal(K/K). This groupis only defined up to inner automorphism, i.e. equivalence in the 2-categoryGrp.

(We will fail to do this, but we’ll get some scraps.)

So, Qp is the p-adic numbers, and Qp contains Zp the p-adic integers. Thereis a valuation Qp → Z ∪ {∞} the normalized valation, where v(x) is just thenumber of times p divides x, i.e. v(pnu) = n if u ∈ Z×p .

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K ⊇ OK the integers of K, where OK is a local ring, whose unique maximalideal is pK = (πK) principal.

Therefore, we have v : K× → Z with v(πK) = 1 and v(πnKu) = n, whereu ∈ O×K .1

Say that L/K is an algebraic extension (possibly infinite).

The valuation vK : K× → Z extends to L× → Q, and L ⊇ OL ⊇ pL, whereOL = {0} ∪ {λ ∈ L

∣∣ v(λ) ≥ 0}.The residue field OL/pL = kL = algebraic extension of kK = OK/pK = a finitefield.

If L/K is Galois, then we get a map Gal(L/K) → Gal(kL/kK) and this issurjective (and not injective in general).

Definition. We say that L/K is unramified if the natural map Gal(L/K) →Gal(kL/kK) is an isomorphism.

Here is the set-up: K/Qp is finite, K ⊇ OK ⊇ pK = (πK).

Exercise 2. Show that the following are equivalent:

(a) pL = πKOL.

(b) vK(L×) = Z.

(c) L/K is unramified.

Exercise 3. Show that the compositum of two unramified exten-sions of K is unramified.

Therefore, if L/K is algebraic, then there exists a unique maximal unramifiedsubextension, so an M where L ⊇ M ⊇ K, M/K unramified and maximal,and Gal(M/K) = Gal(kM/kK) is cyclic or pro-cyclic.

1Reference: Serre’s Local Fields.

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3 Lecture Three

3.1 Inertia subgroup

Recall from last time: We had K/Qp a finite extension. Let L/K be an algebraic,normal extension (hence Galois).

The Galois group Gal(L/K) surjects onto the Galois group Gal(kL/kK). We canthen define the inertia subgroup IL/K to be the kernel of the map Gal(L/K) →Gal(kL/kK). Since kK is finite, we see that the quotient of Gal(L/K) by IL/K isprocyclic (topologically generated by one element).

To understand Gal(L/K), we wish to focus on IL/K .(In theory we don’t lose much by pretending in our heads that L/K is a finite

extension.)IL/K is a closed subgroup of Gal(L/K) since it is the preimage of a closed set

under a “sufficiently nice” map. By the fundamental theorem of Galois theory, IL/Kcorresponds to a subextension M/K, which is the union of all unramified subexten-sions K ′/K.

An interesting special case follows. When L = K, the maximal unramified subex-tension of K (which we denote Knr) satisfies Gal(Knr/K) ∼= Gal(kK/kK) ∼= Z canon-ically; this is due to the fundamental theorem of Galois theory again.

Consider for example K = Qp; then we will find that

Knr =⋃m≥1p-m

Qp(ζm).

Exercise 4. Is this true for general K?

Now we consider L/K Galois with finite inertia group IL/K . (For example, thisis true when L/K is finite.)

We know that IL/K is a normal subgroup of Gal(L/K). We will put a filtrationon IL/K as follows: if σ ∈ IL/K , we consider σ as a map L→ L.

Exercise 5. Check that σ(OL) = OL and σ(pL) = pL.

Because IL/K is finite, we in fact have pL principal. Consider the maximal un-ramified subextension M ; we note that pM = πKOM (where pK = πKOK) becauseM/K unramified. Then vL : L→ Z the discrete valuation satisfies vL = (#IL/K)vKon K×.

Let πL be the appropriate uniformizer. If i ≥ 1, we define IL/K,i = {σ ∈ IL/K :σ(πL)/πL ∈ 1 + piL}.

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Set IL/K,0 = IL/K .

Exercise 6. Check that each IL/K,i is a subgroup of IL/K and that eachis a normal subgroup of Gal(L/K).

Furthermore, if σ is not the identity, σ(πL)/πL 6= 1. Hence for i� 0 we see thatIL/K,i = {1}.

We note that IL/K/IL/K,1 ↪→ k×L via σ 7→ σ(πL)/πL. In particular, IL/K/IL/K,1 iscyclic of order prime to p.

Exercise 7. If i ≥ 1, show that

IL/K,i/IL/K,i+1 ↪→ piL/pi+1L

via σ 7→ σ(πL)/πL − 1, as abelian groups. (Recall that piL/pi+1L is

isomorphic to the additive group of kL).

Exercise 8. In particular, if i ≥ 1, show that IL/K,i/IL/K,i+1∼= (Z/pZ)ni

for some ni, so that its order is a power of p and each non-identity el-ement has order p.

The upshot is that IL/K,1 is the unique Sylow p-subgroup of IL/K , with the quo-tient IL/K/IL/K,1 cyclic of order prime to p. (In particular IL/K is a solvable group.)

We say that L/K is tamely ramified if IL/K,1 is trivial. (Unramified extensionsare tamely ramified.) Otherwise we say L/K is wildly ramified.

We’re really interested in the case L = K, in which case IL/K is not finite. The“lower numbering” IL/K,i does not behave nicely with regard to extensions of L.

This means that if we have extensions L′/L/K with both L,L′ Galois over K andIL′/K finite, we have a natural surjection IL′/K → IL/K , but IL′/K,i is not identifiedwith IL/K,i in general.

However, we can fix this! We introduce a new relabelling of the filtration.Say we have L/K Galois and IL/K finite as before. Let gi = #IL/K,i. Then

g0 ≥ g1 ≥ . . . . Define φ : [0,∞)→ [0,∞) a piecewise linear and continuous function;it is linear on each interval (i, i+1) with φ(0) = 0 and the slope on (i, i+1) is gi+1/g0.

Then eventually φ becomes linear with slope 1/g0. It is clearly a strictly increasingbijection.

Now for any v ∈ R≥0, define IL/K,v = IL/K,dve where d·e is the ceiling function.We can now define the upper numbering. For u ∈ R≥0, define IuL/K =

IL/K,φ−1(u).

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Proposition 1. If L′/L/K are all Galois extensions as usual andIL′/K is finite, then the image of IuL′/K under the obvious map is exactlyIuL/K.

Proof. See Serre’s Local Fields or Cassels-Frohlich (p.38).

The point is that the upper numbering extends to an infinite extension.

Theorem 3.1 (Hasse-Arf). If L/K is abelian, then the jump discon-tinuities in the upper numbering IuL/K are at integers.

Now for any L/K Galois extension, we may define IuL/K by “gluing together”

IuM/K for M/K algebraic with IM/K finite.

3.1.1 Tamely ramified extensions

Recall that when IL/K is finite, we say L/K is tamely ramified with IL/K,1 is trivial -equivalently, IL/K,ε = IδL/K = {1} for any ε, δ > 0. (Using IδL/K works to define tame

ramification for any Galois L/K.)The usual story is true: the compositum of two tame extensions is tame. There-

fore we can talk about maximal tamely ramified subextension.We now have the following general picture. Given L/K Galois, we have a maximal

unramified subextension K1 and a maximal tamely ramified subextension K2, withK1 ⊂ K2 and L/K2 possibly wildly ramified with Gal(L/K2) a pro-p group.

Now consider L = K; we get K/Kt/Knr/K a tower of extensions (Kt is themaximal tamely ramified extension). We now may ask: what is Kt?

If L/K is finite and Galois, consider the tower of extensions L/K2/K1/K asbefore. The Galois group Gal(K2/K1) is IL/K/IL/K,1, which is cyclic of order primeto p.

By Kummer theory (see article by Birch in Cassels-Frohlich), Knr contains allthe mth roots of unity for m prime to p, so if Gal(K2/K

nr) ∼= Z/mZ for m prime top, then K2 is Knr( m

√α) for some α ∈ Knr.

Exercise 9. Check that K2 = Knr( m√πK).

We can now check that

Kt =⋃m≥1p-m

Knr( m√πK).

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We find that Gal(Knr( m√πK)/Knr) is canonically isomorphic to µm via σ 7→

σ( m√πK)/ m

√πK .

Then Gal(Kt/Knr) ∼= lim←−µm where the projective limit is taken over m prime top. This is noncanonically isomorphic to lim←−Z/mZ for m prime to p, or

∏` 6=p Z`.

So then we have a pro-p group Gal(K/Kt), a noncanonical isomorphism Gal(Kt/Knr) ∼=∏6=p Z`, and canonically Gal(Knr/K) ∼= Z.Now we want to understand Gal(Kt/K), which we know as a semidirect product,

though we do not know yet how Z acts on∏

`6=p Z`. Let Frob be the canonical

generator of Gal(Knr/K) ∼= Gal(kKnr/kK) ∼= Z (it is the #kK-power map on theresidue fields).

If we are able to lift Frob to Gal(Kt/K), then it acts by conjugation on thenormal subgroup Gal(Kt/Knr) (since Gal(Kt/Knr) is abelian), and we will be ableto describe Gal(Kt/K).

Recall Gal(Kt/Knr) is canonically isomorphic to lim←−µm(K).

Exercise 10. Check that the map induced by Frob is the q-power mapζ 7→ ζq, where q = #kK.

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4 Lecture Four

We just saw an attempt to analyze the group Gal(K/K) via an explicit attack onthe inertia group. The obstacle is that determining the Sylow-p subgroup of theinertia group is hard. Another approach would be to understand the abelianisationof Gal(K/K).

Let K/Qp be finite. We recall the exact sequence

1→ IK/K → Gal(K/K)→ Z→ 1.

We have Z = 〈Frob〉 topologically, but it has no well-defined lift to Gal(K/K).Frob ∈ Z and we put

(Frob)Z = {. . . ,Frob−1, 1,Frob, . . .} = Z ⊆ Z.

Definition. The Weil Group, WK is formally defined as

WK = {g ∈ Gal(K/K) : Im(g) ∈ Z is in (Frob)Z = Z}.

Pictorially we have the following commutative diagram whose rows are exact:

1 IK/K WK (Frob)Z 1

1 IK/K Gal(K/K) Z 1

The Weil group is a dense subgroup of the Galois group. We give WK a topologywhich is not the subspace topology. We define the topology as follows: IK/K isopen in WK with the usual topology so WK/IK/K = Z with the discrete topology.Consider the following diagram

WK Z

Gal(K/K) Z

Since Z has discrete topology, Z→ Z is continuous. Z has profinite topology and themap Gal(K/K)→ Z is also continuous. WK is the pull-back of the Galois group in

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the category of topological groups. WK surjects onto Gal(L/K) for finite extensionsL/K; this follows from the observations that

Z/nZ ' Z/nZ and Z � Z/nZ.

Define for G a topological group, the subgroup Gc to be the topological closure ofthe normal subgroup of G generated by ghg−1h−1 for g, h ∈ G. Then Gab = G/Gc isthe maximal abelian Hausdorff quotient of G.

Theorem 4.1. (Main theorem of Local Class Field Theory) Let K/Qp

be a finite extension. Then there is a canonical isomorphism (calledthe Artin map)

rk : K× → W ab

K .

(For details refer to Serre’s article in Cassels-Frohlich). We now list out the propertiesof this Artin map:

1. rK(OK) = Im(IK/K).

2. rK(1 + piK) = Im(I iK/K

).

3. rK(πK) ∈ Frob−1 · Im(K/K).

Remark. If X, Y are two abelian groups and ϕ : X → Y is a canonicalisomorphism then ψ : X → Y defined by ψ(x) = ϕ(x)−1 is just as canonical.So there are two canonical isomorphisms between K

×and W ab

K . These are rKand rK ◦ (x 7→ x−1). The way to tell them apart is by the last condition; theone we use sends πK ∈ K

×to the inverse of the Frob.

As per Deligne’s definition we call Frob−1 the geometric Frobenius and Frobthe arithmetic Frobenius.

4. If L/K is finite then Gal(K/L) ↪→ Gal(K/K). Thus there is a canonicalinjection of the Weil groups. We have the following commutative diagram(where we note that the bottom arrow need not be injective)

WL WK

W abL W ab

K

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5. The following diagram commutes:

L×

W abL

K×

W abK

NL/K

rL

rK

Since we know that K×↪→ L

×there must be a map W ab

K → W abL . There exists

the Verlagerung/Transfer map: let H ≤ G of finite index. There exists

V : Gab → Hab; g 7→∏i

γigγ−1i

where {γi} is the set of coset representatives of H in G. Then we have thefollowing commutative diagram

L×

W abL

K×

W abK

rL

rK

Transfer

6. Local class field theory tells about cohomology groups. Let L/K be Galois andfinite, then WL is a normal subgroup of WK (of finite index) and WK/WL =Gal(L/K). In fact, W c

L ≤ WL ≤ WK and W cL is a characteristic subgroup and

hence normal in WK . We define

WL/K = WK/WcL

and this gives us the exact sequence

1→ L×

= W abL ↪→ WL/K → Gal(L/K)→ 1.

This gives rise to an element of H2(Gal(L/K), L×

) called αL/K such that

H2(Gal(L/K), L×

) = 〈αL/K〉 is cyclic of order n = [L : K]. This elementis called the fundamental class.

Remark. There are lots of cohomology groups one can compute using thefundamental class, using the cup product (for example).

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Upshot: We now understand Galois groups of maximal abelian extensions of K.

Exercise 11. Show that the compositum of two abelian extensions isabelian, and therefore that there exists a maximal abelian extensionKab ⊆ K. Show that Gal(Kab/K) = Gal(K/K)ab.

We have the following commutative diagram:

IKab/K Gal(Kab/K) Gal(Knr/K) = Z

IKab/K W abK ' K

× Z

The isomorphism Gal(Kab/K) ' O×K × Z is non-canonical, as it depends on thechoice of Frob.

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5 Lecture Five

Today we will work towards the statement of the local Langlands conjecture. Vaguelyspeaking, the local Langlands conjecture for GLn over K (where K/Qp is a finiteextension) is that there is a one-to-one correspondence between certain n-dimensionalrepresentations of a particular Galois group, and representations of GLn(K).

Let us consider the first of these sets. The Galois group in which we are interestedis a “Weil-Deligne group” (related to the Weil group WK), and so we are motivatedto begin investigating representations of WK .

Recall the definition of WK . We have the following commutative diagram whoserows are exact:

1 IK/K WK FrobZ 1

1 IK/K Gal(K/K) Z 1

id

Now, let E be a field, and equip both E and GLn(E) (for some fixed n ≥ 1) withthe discrete topology. We consider what it means for a homomorphism of groupsρ : WK → GLn(E) to be continuous: we have that ρ will be continuous if and only ifker ρ is open in WK . Because IK/K is compact in WK and ρ is continuous, we knowthat ρ(IK/K) is a compact subset of GLn(E), which is therefore finite.

Thus we can use the theory of lower numbering on ρ(IK/K): because IK/K =

Gal(K/Knr), we deduce the existence of a subextension Knr ⊆ L = L(ρ) ⊆ K suchthat Gal(L/Knr) = ρ(IK/K). Thus

ρ(IK/K) = IL/K ⊇ IL/K,1 ⊇ · · ·

Now, define f(ρ) to be the conductor of ρ, i.e.

f(ρ) =∞∑i=0

1

[IL/K : IL/K,i]dim(V/V IL/K,i),

whereρ : WK → GLn(F ) = AutE(V ), V = En,

and V H is the set of points of V fixed by H for H ≤ Aut(V ).Note that our sum is finite, as for i� 0 we know IL/K,i = 1 and so the dimension

of the quotients vanishes for large enough i.

18

Exercise 12. Show that f(ρ) = 0 if and only if ρ is unramified, if andonly if ρ(IK/K) = 1. Show that f(ρ) is an integer.

Note also that

f(ρ) = dim(V/V IL/K ) +∑i≥1

1

[IL/K : IL/K,i]dim(V/V IL/K,i)

For example, recall the isomorphism rK : K× → W abK from before; this, together

with the surjection WK → W abK gives us another surjection WK → K×. In addition

to the normalized valuation vK : K× → Z, we also have the notion of a norm on K,given by

‖λ‖ = εvK(λ),

where 0 < ε < 1. Two norms ‖ · ‖1, ‖ · ‖2 are called equivalent if ‖x‖1 < 1 if andonly if ‖x‖2 < 1 for every x ∈ L; with our norm, any different choice ε will yield anequivalent norm.

Now, if K is any field which is complete with respect to a non-trivial, non-archimedean norm, we can build the theory of rigid geometry over K. For instance,let K = C((t)) be the field of formal Laurent series in t over C, and let

v

( ∞∑n=−m

antn

)= −m,

if a−m 6= 0. We put a norm on K by fixing some ε, 0 < ε < 1 and putting ‖f‖ = εv(f).Here, there is no real “canonical” choice of ε. In our situation, however, where K/Qp

is finite, a canonical choice of ε is given by ε = q−1, where q = |kK |.Indeed, because kK is finite, we know that K is locally compact, and indeed a

compact open neighbourhood of the identity is given by OK ∼= Zdp (for some d ≥ 1).As such, there exists an additive Haar measure µ on K. Because OK is compact,we have that µ(OK) is finite, so we will scale µ so that µ(OK) = 1. Then what isµ(pK)?

Recall that a Haar measure on a locally compact abelian group G is anytranslation-invariant measure, and furthermore that all Haar measures on G areproportional. Clearly

µ(X∐

Y)

= µ(X) + µ(Y )

if X, Y ⊆ G are compact; therefore, since

OK =∐λ∈kK

λ+ pK

19

(where λ is any lift of λ to OK), we have µ(OK) = qµ(pK); thus µ(pK) = q−1.Now, if a ∈ K×, then define ‖a‖ to be the factor by which multiplication by a

scales the Haar measure; for instance, if K = R, let X = [0, 1], µ(X) = 1 and chooseλ ∈ R×. Then

µ(λX) = µ([0, λ]

)= |λ|µ(X),

thus ‖λ‖ = |λ| (i.e. the usual absolute value).

Exercise 13. Compute the norm ‖z‖ for z ∈ C, using this method.(Hint: for compact X with nonzero measure, we must have ‖λ‖ =µ(λX)/µ(X)).

Back to the case of K a finite extension of Qp: we have our normalized valuationv so that v(πK) = 1, we want ‖πK‖ to equal simply our choice of ε; does there exista canonical choice? Indeed, we have µ(OK) = 1, thus

‖πK‖ = µ(πKOK) = µ(pK) = q−1,

so we have a canonical choice in ε = q−1, as claimed. That is, there is a natural normon K

‖ · ‖ : K× → R>0, ‖λ‖ = q−vK(λ),

and ‖0‖ = 0. In fact, if vK is normalized, then ‖ · ‖ takes only rational values,and a group homomorphism K× → Q>0 is obtained. Thus we have the followingcommutative diagram:

WK W abK K×

Q>0

rK

‖ · ‖‖ · ‖

This defines a map ‖ · ‖ : WK → Q>0.For example: if E = Q, then ‖ · ‖ : WK → GL1(E) is a representation; more

generally, all integer powers of ‖ · ‖ will yield a representation of WK .

For another example, suppose f(‖ · ‖m) = 0; what is ‖Frob‖? (As before, Frob ∈WK is a lift of Frob ∈ Z). Then for the above diagram to commute, we must have

that rK(πK) = Frob−1(IKab/K) maps to q−1, and hence ‖Frob‖ = q.Now, we introduce the notion of a Weil-Deligne representation. This is a

pair (ρ0, N), where ρ0 : WK → AutE(V ) ∼= GLn(E) is a continuous representation

20

as before (E is a field of characteristic 0 with the discrete topology, and V = En),and N is a nilpotent endomorphism of V such that, for all σ ∈ WK , one has

ρ0(σ)Nρ0(σ)−1 = ‖σ‖N. (∗)

For a trivial example: let ρ0 be any continuous representation, and N = 0. Thusevery representation gives rise to a Weil-Deligne representation.

For a nontrivial example: let E = Q and V = 〈e1, e0〉Q. Let

ρ0(σ) =

(‖σ‖ 0

0 1

), N =

(0 10 0

)so that

ρ0(σ)e1 = ‖σ‖e1, ρ0(σ)e0 = e0,

andNe0 = e1, Ne1 = 0.

We check that property (∗) holds:(‖σ‖ 0

0 1

)(0 10 0

)(‖σ‖−1 0

0 1

)=

(0 ‖σ‖0 0

)= ‖σ‖N.

Exercise 14. Check that property (∗) holds when V = Qn with basise0, . . . , en−1 with representation ρ0(σ)ei = ‖σ‖iei and nilpotent operatorNei = i+ 1, i = 0, . . . , n− 2, Nen−1 = 0.

We say that a Weil-Deligne representation (ρ0, N) is F -semisimple if ρ0(Frob)is a semisimple matrix (i.e. diagonalizable over E); this notion is independent of

choice of lift Frob.

Remark. In general, we call a representation of a group G is semisimple if it is adirect sum of simple representations. With a little work, one can show that in thecase G = WK , and ρ0 is continuous, (ρ0, N) is F -semisimple if and only if ρ0(G) issemisimple.

With this machinery, we can now define one half of the local Langlands corre-spondence for GLn: it will be the set of isomorphism classes of all n-dimensional,F -semisimple Weil-Deligne representations of WK . We will define the other half inthe next lecture.

21

6 Lecture Six

6.1 Representations of GLn(K)

In stark contrast to previous lectures, almost all representations here will be infinite-dimensional.

Here’s the set-up. Let E be a field, discretely topologized. Let V be an E-vectorspace (possibly infinite-dimensional). Let K/Qp finite.

Let π : GLn(K)→ AutE(V ) be a group homomorphism.There are too many π’s. We want a sensible notion of continuity.

Definition. Say that π is smooth if for all v ∈ V , the stabilizer Stabπ(v)df= {g ∈

GLn(K)∣∣ π(g)(v) = v} is open. (We have put the p-adic topology on GLn(K).)

Remark. This is equivalent to the map GLn(V ) × V → V being continuous. Onone hand, if the stabilizer is open, then the preimage of any v along the action splitsinto its cosets; on the other hand, if the preimage of v along the action is open, thenyou can intersect it with the open set GLn(V )×{v} to obtain Stabπ(v)×{v}; sincethe product topology is the box topology for a finite product, Stabπ(v) is open.

Definition. Say that a smooth π is smooth-admissible or just plain admissible, iffor all U ⊆ GLn(K) an open subgroup, the fixed points V U of U is finite-dimensional.

If we say “π-admissible”, we mean that π is smooth-admissible.For example, if the dimension of V is 1, then the trivial representation π(g) = 1

for all g is smooth and admissible. If the dimension of V is infinite, then the trivialrepresentation is smooth, but not admissible.

This is tricky to show, but true: π irreducible and smooth implies π admissible.2

Let’s describe the topology on GLn(K).

Definition. A basis of open neighborhoods of the identity in GLn(K) is given bythe matrices {

M ∈ GLn(OK)∣∣M ≡ In mod pmk

}m∈Z≥0

.

Remark. Recall that π is called irreducible if there exist exactly two GLn(K)-invariant subspaces, namely 0 and V .

Here’s a vague statement of the local Langlands conjectures for GLn.

2Question: why? Who proved this?

22

• Local class field theory tells us that K×∼−→ W ab

K .

But we want to understand all of WK . Langland’s insight was that you couldreinterpret this local class field theory isomorphism: if you see the abelianiza-tion of a group, this tells you something about its 1-dimensional representationtheory. Therefore, irreducible 1-dimensional representations of K× should bethe same as irreducible 1-dimensional representations of WK , and the trickis that we’re going to study the rest of WK by studying the rest (i.e. thehigher-dimensional representation theory of K×.)

• Local Landlands conjecture for GLn. There exists a canonical (functorial,definable, bi-interpretable, etc...) bijection

irreducible, smooth-admissible representa-tions of GLn(K)

{n-dimensional, F -semi-simple Weil-Deligne rep-resentations of WK .

}

Remark. Of course, we have to be careful about the criteria by which we define“canonical”. It’s easy to check that either side of this correspondence has the saecardinality, for example, but any randomly chosen bijection won’t do.

Next time, we’re going to convince ourselves is that when n = 1, this is preciselylocal class field theory, and then we’re going to see lots of examples in the case ofn = 2 of elements on both sides.

23

7 Lecture Seven

7.1 Some comments

Last time we finished by stating the local Langlands correspondence for GLn. Todaywe’re going to think about E = C-representations (on both sides of the correspon-dence) for concreteness.

We can interpret the word ‘canonical’ in the statement of the LLC as meaningthat the bijection satisfies a long list of nice properties – e.g. there is a notion ofduality on both sides, we may define L-functions and ε-factors, etc.

A historical note: the list of ‘nice properties’ for the bijection in the LLC forGLn became sufficiently long that it became a theorem that there was at most onebijection satisfying the nice properties. It turns out that there is at least one suchbijection satisfying these properties. In the function field case, the existence of thisbijection is a theorem of Laumon, Rapoport, and Stuhler; the p-adic field case is atheorem of Harris and Taylor. The proofs in both cases are global.

Two obvious observations:

1. This is a brilliant generalization of local class field theory. Given a p-adic fieldK we have to choose an algebraic closure K, and it is the representations ofGal(K/K) that turn out to be important.

2. This is completely pointless as it relates two uninteresting sets, since we haveseen neither Weil-Deligne representations nor smooth admissible irreduciblerepresentations of GLn(K) showing up elsewhere in mathematics. We’re insome sense bijecting two rather pathological collections of objects.

Today we will begin to check that the local Langlands correspondence is useful.

Remark. For any connected reductive group G over K (meaning, roughly, over Kour group G is isomorphic to some nice group such as GLn, O(n), Sp(n), etc.), wecan formulate a local Langlands correspondence for G between smooth irreducibleadmissible representations of G(C) and certain Weil-Deligne representations of someWeil group associated to the L-group of G(C). However, this is not a bijection but asurjection with finite fibers, which again satisfies some long list of natural properties(though in this case the list does not uniquely characterize the map). The finitefibers of the map are called ‘L-packets’.

24

7.2 Local Langlands for GL1

One side of local Langlands for n = 1 consists of 1-dimensional Weil-Deligne repre-sentations (ρ0, N) : WK → GL1(C) (note that we must have N = 0 and ker ρ0 closed- hence ρ0 factors through W ab

K∼= K×). Hence this side of the correspondence is just

1-dimensional continuous complex representations of W abK .

The other side of the correspondence consists of smooth irreducible admissiblerepresentations of K×. It is not difficult to check that an admissible irreduciblerepresentation of K× must have finite dimension - in fact, it must be one-dimensional.Furthermore continuity turns out to be equivalent to smoothness. Hence this side ofthe correspondence is just continuous group homomorphisms K× → C×.

Then rK of local class field theory gives us the desired canonical bijection.If π : K× → C× is a smooth admissible irreducible one-dimensional represen-

tation, we may define the conductor f(π) to be 0 if π|O×K is trivial and otherwise

f(π) = r, for r the smallest positive integer so π|1+PrKOK is trivial.One can check that the conductors of ρ0 and the corresponding π are the same.

7.3 Weil-Deligne representations arising from `-adic repre-sentations

As ever, K/Qp is a finite extension. Say ρ : Gal(K/K) → GLn(Q`) is a continuousrepresentation of the absolute Galois group of K (GLn(Q`) carries the `-adic topologyfor ` 6= p a prime).

The wild part of the inertia subgroup will end up being finite under this repre-sentation, but the same cannot be said of the tame part.

These `-adic representations show up in ‘nature’: for example, the `-adic Tatemodule of an elliptic curveE/K, or more generally, `-adic etale cohomologyH i

et(XK ,Q`)of an algebraic variety X/K. Furthermore `-adic deformations of examples give newexamples.

Remark. If E/K is an elliptic curve with split multiplicative reduction, then E(K) ∼=K×/(qZ) for some q ∈ K, |q| < 1 canonically. This allows one to explicitly compute

the `-adic Tate module of E as T`E/IK/K =(cyclo ∗

0 Id

)where ∗ can be nontrivial and

infinite.

Recall that ` 6= p, so if ρ is an `-adic representation as above, ρ(IK/K) can beinfinite, but ρ(Iε

K/K) is finite for ε > 0 since Iε

K/Kis a pro-p-group. Also recall that

Gal(Kt/Knr) ∼=∏

r primer 6=p

Zr,

25

so the only part we need to worry about is the Z`-part.Fix t : Gal(Kt/Knr) → Z` a surjection. Also fix φ ∈ Gal(K/K) lifting Frob ∈

Gal(Knr/K).

Proposition 2 (Grothendieck). If ρ : Gal(K/K)→ GLn(E) for E =Q` is a continuous `-adic representation, then there exists a uniqueWeil-Deligne representation (ρ0, N) : WK → GLn(E) such that ρ(φmσ)for any σ ∈ IK/K ,m ∈ Z is equal to ρ0(φmσ) · exp(N.t(σ)). In fact theisomorphism class of (ρ0, N) is independent of the choice of φ, t.

Note that taking ρ0 to be the restriction of ρ would not necessarily work due tothe fact ρ is continuous with respect to the `-adic topology on E while ρ0 must becontinuous with respect to the discrete topology on E. We may define the exponentialof a nilpotent matrix via the usual power series.

Remark. In the Tate curve example, it turns out that N is nonzero; indeed, N =( 0 1

0 0 ).

Remark. Not all Weil-Deligne representations arise in this way, since those (ρ0, N)that do must have that all eigenvalues of ρ0(φ) be `-adic units. However, this isin some sense the only obstruction - if the eigenvalues of ρ0(φ) are `-adic units forsome Weil-Deligne representation (ρ0, N), then there exists ρ so the Weil-Delignerepresentation arises from ρ.

7.4 Smooth admissible representations of GL2(K)

Say we have χ1, χ2 : K× → C× continuous admissible characters. Define I(χ1, χ2) tobe the vector space of functions φ : GL2(K)→ C (pointwise scalar multiplication andaddition are the operations) such that φ is locally constant for the p-adic topologyand φ (( a b0 d ) g) = χ1(a)χ2(d)‖a/d‖1/2φ(g).

This can also be described as an induced representation from the upper triangularsubgroup B to GL2(K).

Define π : GL2(K)→ AutC(I(χ1, χ2)) by (π(g)φ)(h) = φ(hg) for g, h ∈ GL2(K), φ ∈I(χ1, χ2).

26

8 Lecture Eight

Lemma 1. Let B(K) be the upper triangular matrices. Then

GL2(K) = B(K) ·GL2(OK) = {bg : b ∈ B(K), g ∈ GL2(OK)}.

Proof. Let M =

(a bc d

)∈ GL2(K). We want β ∈ B(K) and γ ∈ GL2(OK) such

that M = βγ.

Without loss of generality, we may assume M ∈ SL2(K) as we can left multiply by(det(M)−1 0

0 1

). We may further assume c, d ∈ OK and at least one of them is a

unit. This is because we can choose α ∈ K× such that αc, αd ∈ OK and at least one

is a unit by left multiplication by

(α−1 00 α

). If c is not a unit, we can multiply on

the right by

(0 −11 0

)∈ GL2(OK). We can switch c, d so without loss of generality

we assume that c is a unit in OK . We see(1 −a/c0 1

)(a bc d

)=

(0 −1/cc d

)∈ GL2(OK).

Remark. Let ϕ : GL2(K) → C be locally constant; then ϕ is continuous withrespect to the discrete topology on C and hence its image must be finite. ϕ(B(K))is controlled by the definition of I(χ1, χ2).

Recall: I(χ1, χ2) is a vector space, and is smooth and admissible. We have the norm‖ · ‖ : K

× → Q≥0, defined by πK 7→ q−1K .

Question: Is I(χ1, χ2) irreducible?No, this is easy to see if we study Inaive(χ1, χ2) = {ϕ(Mg) = χ1(a)χ2(d)ϕ(g)}, where

M =

(a bc d

). and noticing that Inaive(χ1, χ2) = I(χ1‖ · ‖−1/2, χ2‖ · ‖1/2).

If χ1 = χ2 are both the trivial representation, using Frobenius reciprocity we cansee that Inaive(χ1, χ2) contains the constant functions, and therefore contains a one-dimensional G-invariant subspace. More generally, we can show that if χ1 = χ2 thenInaive contains a one-dimensional G-invariant subspace. There is another case when

27

I(χ1, χ2) is not irreducible, this is when χ1/χ2 = ‖ · ‖±2. In this case Inaive(χ1, χ2)has a one-dimensional quotient.

This is what happened: There’s a duality, i.e., there is a natural pairing

I(χ1, χ2)× I(χ−1, χ−12 )→ C

involving an integral on G and B. At some point we change left Haar measure onB to a right Haar measure. These don’t coincide and differ by the “fudge factor”‖a/d‖1/2. Thus, the dual is I(χ−1

1 , χ−12 ).

Remark. (Theorem 1.21 Bernstein-Zelevinsky) Turns out I(χ1, χ2) is irreducible ifχ1/χ2 6= ‖ · ‖±1. If χ1/χ2 = ‖ · ‖−1 then we have the exact sequence

0→ ρ→ I(χ1, χ2)→ S(χ1, χ2)→ 0

where ρ = (ρ, V ) is the one-dimensional representation of GL2(K) given by g 7→(χ1×‖ · ‖1/2

K )(det g). It is a fact that S(χ1, χ2) is irreducible. It follows from dualitythat if χ1/χ2 = ‖ · ‖1 then there is an exact sequence

0→ S(χ2, χ1)→ I(χ1, χ2)→ ρ→ 0

There is a completely different construction (cf. Jacquet-Langlands; for understand-ing the articles in this book, Bump’s book is helpful). It was an observation of Weilthat for any field k one can write a precise presentation of SL2(k):⟨A(t) =

(t 00 t−1

), B(u) =

(1 u0 1

),W =

(0 −11 0

) ∣∣∣∣ explicit obvious relations

⟩Upshot: We can construct representations of SL2(k) by giving explicit actions of thegenerators on some C-vector space, and then checking relations. Weil observed thatwe can use the vector space of L2 functions on k. These are another huge source ofrepresentations of SL2(k), and hence GL2(k).

(Theorem 4.6 in J-L, page 72) Let L/K be a quadratic extension. If χ : L× → C×

is admissible, with χ 6= χ ◦ σ (1 6= σ ∈ Gal(L/K)), then one can construct anirreducible, infinite-dimensional representation BCK

L (χ) of GL2(K) on the space ofsquare-integrable functions on L.

Remark. 1. I(χ1, χ2), S(χ, χ×‖·‖),BCKL (ψ) are all infinite-dimensional, smooth-

admissible representations. The last two are always irreducible, and the firstis irreducible for the cases mentioned in the above discussion.

28

2. The only isomorphisms between the representations I(χ1, χ2) have the formI(χ1, χ2) = I(χ2, χ1), when χ1/χ2 6= ‖ · ‖±1.

3. When char(κK) = p > 2, then these are all the infinite-dimensional irreduciblesmooth-admissible representations of GL2(K). This is not true for p = 2.

4. The only finite-dimensional representations π of GL2(K) are one-dimensional,and hence factor through the abelianisation. Thus, it’s of the form χ ◦ detwhere χ : K

× → C× .

29

9 Lecture Nine

Recall the examples we have seen so far of smooth, irreducible, admissible represen-tations of GL2(K), namely:

• I(χ1, χ2), when χ1/χ2 6= ‖ · ‖±1;

• S(χ, χ× ‖ · ‖);

• χ ◦ det; and

• BCKL (ψ).

It is a fact that, if p = char(kK) > 2, these are the only smooth irreducible admissiblerepresentations of GL2(K).

If G is a connected, reductive group over K and π is a smooth admissible repre-sentation of G(K), then there is a notion of “genericness” of π: if G = GL2, then πis called generic if dim(π) is infinite.

Hence, suppose π is an irreducible, admissible representation of GL2(K), andassume that π is infinite-dimensional.

Theorem 9.1. (Casselman) For n ≥ 0, define

U1(pnK) =

{(a bc d

)∈ GL2(OK) : c ≡ 0 mod pnK , d ≡ 1 mod pnK

}.

Then every U1(pnK) is compact-open, and the quantity d(π, n) := dim(πU1(πnK)

)is finite. Furthermore,

d(π, n) = max(0, 1 + n− f(π)

)for all n ≥ 0.

Proof. Omitted (see Casselman’s Antwerp conference proceedings).

Exercise 15. Check that the theorem holds for I(χ1, χ2), with χ1/χ2 6=‖ · ‖±1.

Exercise 16. Check that

f(I(χ1, χ2)

)= f(χ1) + f(χ2)

and that

f(S(χ, χ× ‖ · ‖

)=

{1 if f(χ) = 0,

2f(χ) iff(χ) > 0.

30

Exercise 17. Prove Schur’s lemma: if π is an irreducible admissiblerepresentation of GLn(K), then there exists an admissible characterχπ : K× → C× such that, for all λ ∈ Z

(GLn(K)

) ∼= K×, the actionof λ on π is that of the scalar χπ(λ); the character χπ is called thecentral character of π.

Exercise 18. Show that

χI(χ1,χ2) = χ1χ2 = χS(χ1,χ2) and χφ◦det = φ2

for any φ : K× → C×.

Now, returning again to the local Langlands correspondence, we introduce somenotation: assuming that the correspondence is true for GL1, write χi for the characterK× → C× associated to the representation ρi : WK → C× for i ≥ 1, and conversely.Then the correspondence for GL2 associates:

1. I(χ1, χ2) to the Weil-Deligne representation (ρ0 = ρ1 ⊕ ρ2, N = 0),

2. S(χ1, χ1 × ‖ · ‖) to (ρ0 =

(‖ · ‖ρ1 0

0 ρ1

), N =

(0 10 0

)),

3. χ1 ◦ det to (ρ0 =

(ρ1 × ‖ · ‖1/2 0

0 ρ1 × ‖ · ‖−1/2

), N = 0

), and

4. BCKL (ψ) to

(ρ0 = IndWK

WLσ,N = 0

), where IndGH is the induced representation,

ψ : L× → C× and 1 6= σ ∈ Gal(L/K).

In the case p = char(kK) = 2, a problem arises from the fact that there is a Galoisextension K/Q2 with Galois group isomorphic to S4, which has no abelian subgroupof index 2.

For GL2, the local Langlands correspondence was proven by explicitly writingdown these associations; for general GLn, representation-theoretic techniques areneeded to reduce the problem to matching irreducible pairs (ρ0, N) with so-calledsupercuspidal representations π, using a global argument.

We now introduce some terminology regarding Weil-Deligne representations (ρ0, N).Define the conductor of (ρ0, N) to be

f(ρ0, N) = f(ρ0) + dim(V IK/K/(kerN)IK/K

).

Note that f(ρ0, 0) = f(ρ0) and that det(ρ0) : WK → C×.

31

Exercise 19. Check that if π corresponds to the Weil-Deligne repre-sentation (for the GL2 Langlands correspondence), then f(π) = f(ρ0, N)and that χπ corresponds to det(ρ0) via local class field theory.

Exercise 20. Check that, if p > 2, then the only F -semisimple, two-dimensional Weil-Deligne representations of WK are the ones we havelisted above. (Hint: look up result 2.2.5.2 in Tate’s “Number theoreticbackground”).

We will close our discussion of the local Langlands correspondence with an explicitconsideration of the “unramified” case (i.e., when f(π) = 0) in dimension two. If(ρ0, N) is the corresponding Weil-Deligne representation, then f(ρ0, N) = 0 also, andthere are precisely two options: either

1. π = I(χ1, χ2) for some χi : K× → C× factoring through K×/O×K with χ1/χ2 6=‖ · ‖±1; or

2. π = χ ◦ det for some χ : K× → C× factoring through K×/O×K .

If dimπ =∞, so that π = I(χ1, χ2) as in case 1, then the fact that f(π) = 0 impliesthat πGL2(OK) is one-dimensional.

More generally, if G is a connected, reductive, unramified group over K (such asGLn), and π is a smooth admissible irreducible representation of G(K), then we saythat π is unramified if there exists a hyperspecial maximal compact subgroup H ofG(K) such that πH 6= 0 (such as H = GLn(OK)).

If we want to perform calculations with π, a good place to start is with theinvariant subspace πGL2(OK), which we note is not GL2(K)-invariant. We will do thisby introducing the Hecke operators.

Let G = GL2(K); or, more generally, any locally compact, totally disconnectedtopological group (maybe satisfying some additional properties). Then if π is arepresentation of G and U, V ⊆ G are compact open subgroups (for instance, U =U1(pnK) or GL2(OK)) and g ∈ G, there exists a Hecke operator[

UgV]

: πV → πU ,

which is C-linear, and is defined as follows: write the (compact) subset UgV as adisjoint union of its cosets

UgV =r∐i=1

giV,

32

and put [UgV

]x :=

r∑i=1

gix.

This may be thought of as a sort of averaging process.

Exercise 21. Check that[UgV

]x does indeed lie in πU , and is inde-

pendent of choice of coset representatives gi.

Returning to the GL2(K) case, and assuming f(π) = 0, let us take U = V =GL2(OK), and define the operators S, T to be

S :=[U(πK 00 πK

)V]

: GL1(OK)→ GL1(OK).

andT :=

[U(πK 00 1

)V]

: GL2(OK)→ GL2(OK).

Then dim πGL1(OK) = 1, so by Schur’s lemma, T acts via a scalar t ∈ C and s ∈ C.

Exercise 22. If π = I(χ1, χ2), χ1/χ2 6= ‖ · ‖±1, and f(π) = 0, thenshow that

t =√qK × (α + β) and s = χπ(πK) = αβ,

where α = χ1(πK), β = χ2(πK), and qK = |kK |. Also check the casewhen π is one-dimensional and unramified.

Exercise 23. Show that if π is an irreducible admissible representationof GL2(K) and f(π) = 0 (where π is I(χ1, χ2) or is one-dimensional,then π corresponds to the Weil-Deligne representation (ρ0, 0), where

ρ0 : WK → WK/IK/K∼= Z→ GL2(C),

and ρ0(Frob) has characteristic polynomial X2− t√qKX+s and Frob ∈

WK/IK/K.

33

Exercise 24. If G = GLn and

Ti = GLn(OK)

πKπK

. . .

πK1

. . .

1

GLn(OK)

(with i copies of πK on the diagonal), corresponding to the scalars ti,then compute the characteristic polynomial of ρ0(Frob).

If G = G(K) with G/K unramified, and if π is an unramified representation of G,then Langlands reinterpreted the Satake isomorphism to associate to π a semisimpleconjugacy class in LG(C); in particular, the conjugacy class is that of ρ0(Frob), whereπ corresponds to (ρ0, N).

34

10 Lecture Ten

Part 2: The global Langlands correspondence.In this part, K will be a number field, i.e. a finite extension of Q.We’ll start by talking about the structure Gal(L/K) of a finite Galois extension

L/K, and in particular its relationship to local Galois groups. Taking limits, we getstructure on Gal(K/K). In the local situation, the Weil group wasn’t quite the rightanswer because we needed Weil-Deligne representations; in the global case, we needthe “global Langlands group”, but no one knows what that is...3

However, we still have `-adic representations of Gal(K/K), and these are maybea working definition of the ρ-side of the correspondence. And then on the π-side iswhere we see automorphic representations.

The global Langlands philosophy is that every automorphic representation ofGL2(K) should yield a 2-dimensional representation of the global Langlands group(whatever this is...)

“Uncheckable conjecture”: all automorphic representations for π should corre-spond to n-dimensional representations of a global Langlands group.

Here’s a checkable conjecture instead: all algebraic automorphic representations πof GLn(K) should correspond to “motivic” n-dimensional representation of Gal(K/K)→GLn(Q`) (“unramified outside a finite set, de Rham...”) and these should correspondto “motives”:

{algebraic automorphic rep-resentations π of GLn(K)

} {“motivic” representationsρ : Gal(K/K)→ GLn(Q`)

}

{motives}

Furthermore, if we abelianize the global Langlands group, we should get K×\A×K .In n = 1, the “uncheckable conjecture” will be global class field theory.4

3Quote: ‘You can begin to see why this is called a “philosophy”, because some parts are verywell-defined, and for other parts you have to be... a little bit of a dreamer. I think I read in a paperof Arthur that in 100 or so years time after we’ve proved all the various Langlands correspondences,maybe the very last theorem will be that the global Langlands group exists. Maybe we’ll have toconstruct all the representations and then say, “ha ha! by the Tannakian formalism...” ’

4Quote: ‘I don’t know if this is worth mentioning, but I’ve had this dream—the automorphicrepresentations of GLn(K) should be a subset of a new p-adic space of p-adic automorphic repre-sentations of GLn(K).’

35

10.1 Galois groups

Return to a finite extension K/Q.K contains OK , the algebraic integers in K; for instance if K = Q, then OK = Z.Choose 0 6= p, a prime (and hence maximal). This will be an ideal of OK , with

OK/p =: kp a finite residue field.We can complete K at p. Here’s one way of doing it: we can define

OKp

df= lim←−OK/pn

and we put

Kpdf= Frac

(OKp

).

Alternatively, with p fixed, then if λ ∈ K×, then λOK is a fractional ideal ofK and therefore factors as the product of pvp(λ) with other prime ideals to variouspowers, where vp : K× � Z, and we can define ‖λ‖p a norm on K with ‖0‖p = 0 ad

‖λ‖pdf= (qp)

−vp(λ).The norm on K induces a metric, and now we can complete K with respect to

this metric and get Kp a local field. Kp is a finite extension of Qp, where p ∩ Z = p.Now, let L/K be a finite Galois extension of number fields. We get a finite Galois

group Gal(L/K).Let’s say we’ve got p ⊆ OK as above, and we’ve got this number field L with

pOL⊆L:

OK OL

p pOL

We know from undergraduate abstract algebra that pOL will be an ideal, but maynot be a prime ideal. However, it factors into prime ideals of L, say

pOL = Pe11 Pe2

2 · · ·Pegg ,

with each P a prime ideal of OL.Gal(L/K) acts on L, and on OL (by exercise 5), and acts trivially on K and

hence on p.Therefore, Gal(L/K) maps pOL to pOL.

36

Exercise 25. Show that, if σ ∈ Gal(L/K) then σ(Pi) is a prime idealof OL, so σ(Pi) lies in the set {P1, . . . ,Pg}.5

Exercise 26. Show that the action of Gal(L/K) on the set of primesof OL dividing p is transitive.

As a consequence we see that e1 = · · · = eg. This also means in some sense thatthe Pi are all isomorphic, since they are Galois-conjugate. Therefore,

LP1 ' · · · ' LPg ,

where LP = OL/P.Here’s the set-up. We have L/K finite Galois, p as before, so that pOK factors

into∏

i≤gPeii . Set P = P1, fixed choice of a prime of L.6

DefineDPdf= {σ ∈ Gal(L/K)

∣∣σ(P) ⊆ P}. Then Gal(L/K)/DP = {P1, . . . ,Pg}.Our task is to understand this DP. If σ ∈ DP, then σ : L → L, OL → OL, and

P→ P. Then again by transport de structure, we get σ : LP → LP fixing Kp.It turns out that LP/Kp is Galois and that DP ' Gal(LP/Kp), so Gal(LP/Lp)

lives inside Gal(L/K).(This is in Serre’s Local fields).Recap: we start with Gal(L/K) and we choose some p ⊆ OK . Then we have

to choose a P | pOL, which induces an identification DP ⊆ Gal(L/K), and we canidentify DP with Gal(LP/Kp).

Inside Gal(LP/Kp) is an inertia subgroup. A global fact: if p does not divide thediscriminant of L/K, then this inertia subgroup is trivial.

If LP is an unramified extension ofKp, then there exists FrobP ∈ DP ⊆ Gal(L/K).FrobP is slightly annoying: because it depends not only on p but also on the

choice of P | pOL. Say P′ is another choice.

5Here’s a trivial observation: if X and Y are objects in mathematics, topological spaces orgroups or whatever, and you do some kind of calculation in X which involves taking elements inX and using axioms or whatever, then because this calculation is definable in X, an appropriatenotion of isomorphism will show the analogous calculation in Y to be valid also. This is calledby Deligne, in French, “transport de structure,” and it’s a completely trivial observation. Butsometimes mathematics is about giving the right definitions or thinking about things the right way.It turns out that thinking clearly about this trivial observation in the particular case where Xhappens to be equal to Y but the isomorphism is not the identity map can sometimes really help,e.g. X = Y = L and σ the isomorphism is in Gal(L/K).

6‘Somehow you remember where you were when you learn things. I learned about modules at abar. I was reading Atiyah-Macdonald.’

37

By transitivity of the action of Gal(L/K) on {P1, . . . ,Pg}, there exists a σ ∈Gal(L/K) such that σ(P) = P′, and then by transport de structure, DP′ = σDPσ

−1

and FrobP′ = σ FrobP σ−1.

Upshot from these considerations: you can define Frobp to be the conjugacyclass of FrobP = {FrobP′ such that P′ | pOL}, and this works for all p not dividingthe discriminant of L over K.

38

11 Lecture Eleven

11.1 Decomposition and inertia subgroups

Recall that for L/K a finite Galois extension and p a nonzero prime ideal of OK ,the ideal pOL Is probably not prime. We may factor pOL as

∏ri=1 P

eii , a product of

prime ideals of OL.As we have seen, Gal(L/K) acts on the set of these Pi - in fact, it acts transitively

on this set.Fix P = Pi for some I and set DP/p = {σ ∈ Gal(L/K) : σ(P) = P}. This is the

decomposition group.

Exercise 27. Check that, if σ ∈ DP/p, then σ acts on the comple-tion LP of L at P and fixes Kp, and so σ ∈ Gal(LP/Kp) = DP/p ⊆Gal(L/K).

We can then define again an inertia subroup IP/p = ILP/Kp ⊆ Gal(LP/Kp) =DP/p.

Now we consider a miracle. There is an ideal ∆ = disc(L/K) ⊆ OK , the dis-criminant of L/K, such that p - ∆ if and only if IP/p = {1} for any P lying overp. (The inertia subgroups are independent of the choice of P up to isomorphism, bytransport de structure.)

Thus for all p excepting some finite set S = {p : p | ∆L/K} we get for any P lyingover p a cyclic group DP/p = 〈FrobP〉 = DP/p/IP/p = Gal(kP/kp).

Remark. Say L/K is finite, then what is 〈FrobP〉? To see that, pick P/p, thenthere exists a unique element σ ∈ Gal(L/K), s.t. σ(x) ≡ x#kp for all x ∈ OL. Thenwe denote FrobP to be the element σ and choose 〈FrobP〉 to be the conjugacy classof that element. One could show that the class is independent of the choice of theprime ideal P lying above p.

However we now encounter an issue: FrobP depends on choice of P lying overp, so p determines a conjugacy class of automorphisms FrobP for P|p. By abuse ofnotation, we will write Frobp for that conjugacy class.

Here is another important fact. Every conjugacy class C of Gal(L/K) is equalto Frobp for infinitely many p. This is a consequence of the Chebotarev densitytheorem.

In fact, the density of primes p such that Frobp = C is #C/#Gal(L/K).

39

11.2 Infinite extensions

For K a number field with fixed algebraic closure K, Gal(K/K) is ramified at everyprime of K.

Let S be some finite set of maximal ideals of OK . We want to be able to defineFrobp for p /∈ S.

It turns out that if K/L1/K,K/L2/K are towers of extensions so that Li/Kfinite and unramified outside S, then the same is true of the compositum.

Thus we can defineKS =

⋃L/K finite Galois

unramified outside S

L.

For example, if S is empty, QS = Q. However for general K a number field, evenif S is empty the extension KS/K may be infinite.

Theorem 11.1. Let S be a finite set of primes of K, let d ∈ Z>0, thenthere exists only finite many L/K satisfying

(1). [L : K] ≤ d;

(2). L/K is unramified outside S.

Theorem 11.2. (Chebotarev) Let L/K be a finite Galois extension.Let C ⊂ Gal(L/K) be a conjugacy class. Then there are infinitely manyplaces v of K, unramified in L, s.t. Frobv = C. In fact,

|C||Gal(L/K)|

= Drichlet density of number of such primes

Now consider K = Q, S = {p}. It is immediately apparent that KS containsQ(ζpn) for any n ≥ 1.

For K = Q, N ∈ Z≥1, define S = {p : p | N}; then similarly KS contains Q(ζN).If p /∈ S is a prime, there should be a canonical conjugacy class Frobp in (Z/NZ)×,which is abelian. It is immediately apparent that we should have Frobp = p mod N .This can be seen by considering the action on ζN and the action on the residue field.

Let’s stick to the case now K = Q, S = {p}. It is immediately apparent that KS

contains⋃∞n=1 Q(ζpn) =: Q(ζp∞). Hence Gal(KS/K) surjects onto Gal(Q(ζp∞)/Q) =

Z×p .If r is prime and r 6= p, we want to find Frobr in Z×p : similar considerations

as before make it clear that in each (Z/pnZ)× = Gal(Q(ζpn)/Q) we have Frobr =r mod pn, so Frobr ∈ Z×p is easily defined.

40

In the more general case, we may write Gal(KS/K) as a projective limit oflim←−Gal(L/K) for L/K finite Galois and unramified outside S – in which case forall p /∈ S we get conjugacy classes Frobp,L/K ⊆ Gal(L/K) for each L which ‘gluetogether’ in a nice way to give a conjugacy class Frobp,KS/K ⊆ Gal(KS/K).

The Chebotarev density theorem does not apply to infinite extensions. However,if L/K is an infinite Galois extension unramified outside S, the map p 7→ Frobp for{p /∈ S} has dense image; that is,

⋃p/∈S Frobp ⊆ Gal(L/K) is dense.

As a corollary, if we have some map F : Gal(L/K)→ X continuous and constanton conjugacy classes, we may be able to recover F from the data F (Frobp) for p /∈ S.

11.3 Representations of Galois groups

Theorem 11.3 (Brauer-Nesbitt). Let G be a group and E a field.Recall that a representation ρ : G → GLn(E) is called semisimple if ρis a direct sum of irreducible representations. If we have two semisimplerepresentations ρ1, ρ2 : G → GLn(E) and for any g ∈ G, {ρi(g)}i=1,2

have the same characteristic polynomials, then in fact ρ1∼= ρ2.

Proof. Algebra.

Remark. The above is not true that if we replace GLn(E) with an arbitrary groupand ‘having the same characteristic polynomial’ with being conjugate to one another.

Remark. If char(E) = 0 it suffices that tr(ρ1) = tr(ρ2) (though both must still besemisimple). If n! is invertible, this is true since we can compute the characteristicpolynomial of ρ(g) from tr(ρ(gi)) for i = 1, . . . , n.

Exercise 28. For G = Z/3Z, E = F2, find non-isomorphic ρ1, ρ2

which are semisimple and reducible such that tr(ρ1) = tr(ρ2).

The upshot of all this is that if we have an `-adic representation ρ : Gal(KS/K)→GLn(E) (where E some finite extension of Q`), continuous with regard to the `-adic topology and semisimple (or just irreducible), the characteristic polynomial ofρ(Frobp) for all p /∈ S is a well-defined polynomial Fp ∈ E[x]. IF we know all Fp forp /∈ S, then ρ is uniquely determined by this data.

Now, as an example, let’s do the cyclotomic character.Let K = Q, S = {p}. Let L = Q(ζp∞). Then Gal(L/K) = Z×p ⊆ GL1(Qp).

By the fundamental theorem of Galois theory, Gal(Q/Q) surjects onto Gal(Q(ζp∞)/Q) ⊆GL1(Qp). Hence we get a one-dimensional p-adic representation of Gal(Q/Q). ρ is

41

called the p-adic cyclotomic character, ωp. It is determined by the fact thatρ(Frobr) = r ∈ Z×p for all primes r not equal to p.

A confusing thing: if p, ` are distinct primes, with S = {p, `}, we have tworepresentations ωp : Gal(Q/Q) → Gal(QS/Q) → Z×p and ω` : Gal(Q/Q) → Z×` .Note that Frobr for r /∈ S give a dense subset of Gal(QS/Q).

Then ωp, ω` are two Galois representations that are equal on some dense subset ofGal(QS/Q). Does Brauer-Nesbitt imply ωp = ω`? No, because these representationsare over different fields.

In fact we will find that kerωp, kerω` are very different.

42

12 Lecture Twelve

Another weird example: Let K/Q be a number field and E be an elliptic curve overK, and let S0 be the set of finite places (maximal ideals of OK) where E has badreduction. Let ` be a prime number; the Galois group Gal(K/K) acts on E[`n](K).The Tate module is defined as

lim←−n

E[`n] = T`[E] ' Z` × Z`.

We can realize the Galois action by looking at the map

ρE,L : Gal(K/K)→ GL2(Z`) = Aut(Z` × Z`).

The above map is well defined up to conjugation. ρE,L factors through Gal(KS0∪{p|`}/K).The characteristic polynomial of ρE,`(Frobp) is X2 − apX + N(p) ∈ Q[X] ↪→ Q`[X]where ap = 1 + N(p) − #E(κp) and N(p) is the norm. The trace of ρE,L(Frobp)is ap, which is independent of `. What we need to observe is that ρE,`, ρE,p arenot isomorphic when p 6= `. This is because ρE,` is infinitely ramified at `, butρE,p(wild inertia at `) is finite.

12.1 `-adic Representations

Let K be a number field and E/Q` be a finite extension. Let S be a finite set ofmaximal ideals of OK .

Definition. If ρ : Gal(KS/K) → GLn(E) is continuous with respect to the `-adictopology on the right-hand side, and the profinite topology on the left-hand side,we call ρ an `-adic representation of Gal(K/K). We also say that ρ is unramifiedoutside S.

Definition. We say that ρ is rational over E0 (a subfield of E) if, for all p /∈ S, thecharacteristic polynomial ρ(Frobp) lies in E0[X].

Example 2. 1. Consider the cyclotomic character

ω` : Gal(Q/Q)→ GL1(Q`)

Frobr 7→ r, r 6= `.

Then ω` is rational over Q.

2. Tate modules of elliptic curves are rational over Q.

43

3. (Deligne) `-adic etale cohomology of smooth proper algebraic varieties (overK), H i

et(XK ,Q`) are rational over Q.

Definition. We say that ρ is pure of weight w if ρ is rational over a number fieldE and, for all embeddings i : E ↪→ C and all eigenvalues α of ρ(Frobp) (p /∈ S), wehave

|i(α)| = (#kp)−w/2.

Example 3. 1. (Deligne) H iet(XK ,Q`) is pure of weight i when X is a smooth

proper algebraic variety.

2. Cyclotomic characters are pure of weight −2.

3. Tate modules are pure of weight −1; in this case, the roots of X2−apX+N(p)

are complex conjugates, and |ap| ≤ 2√N(p) (Hasse inequality).

Now ` will vary. Let K be a number field, E0 be a subfield of E, a finite extensionof Q and S0 be a set of finite places. We are given a huge amount of data, as for allp /∈ S0 we have a polynomial Fp ∈ E0[X]. Also, for all maximal ideals λ ⊆ OE0 wehave an `-adic representation

ρλ : Gal(KS0∪{p|`}/K)→ GLn((E0)λ)

where the completions (E0)λ for λ | ` are finite extensions of Q`.

Definition. We say ρλ is a compatible system of λ-adic representations if for all λ | `and p /∈ S0, p - `, the characteristic polynomial ρλ(Frobp) = Fp(X) independent ofλ.

Example 4. 1. For cyclotomic characters, Fp(X) = X−p and so is a compatiblesystem.

2. For Tate modules T`E, for all ` and E0 = Q, one has

Fp(X) = X2 − apX +N(p),

and hence this is a compatible system.

3. H iet are known to be compatible systems.

The main aim of global class field theory is to understand Gal(K/K)ab.

44

12.2 Adeles

Let K be a number field of degree d. There are d embeddings σ : K ↪→ C. Let thenumber of real embeddings be r1 and the number of complex embeddings be 2r2,then d = r1 + 2r2.

Definition. An infinite place of K is either a real place v = σ : K → R or v ={σ, σ ◦ c} : K → C \ R, where c is complex conjugation. We then define

K∞ =∏v

Kv,

the product taken over all infinite places.

Remark. 1. K∞ = K ⊗Q R.

2. K∞ = Rr1 × Cr2 as rings.

3. K×∞ = (R×)r1 × (C×)r2 are not connected in general.

45

13 Lecture Thirteen

Last time we introduced the notion of an infinite place of a number field K. Forinstance, if K = Q

(3√

2), then we have

K∞ = K ⊗Q R =(Q[X]/(X3 − 2)

)⊗Q R = R[X]/(X3 − 2).

The roots of our polynomial are α = 3√

2 ∈ R, w = ζα, and w = ζ2α, where ζ3 is aprimitive third root of unity; thus

K∞ = R[X]/(X − α)(X2 + αX + α2).

The Chinese remainder theorem implies that this is isomporphic to(R[X]/(X − α)

)×(R[X]/(X2 + αX + α2)

) ∼= R× C;

the isomorphism with the real field is canonical, whereas the isomorphism with C isonly unique up to complex conjugation.

Definition. Let K be a number field and p ⊆ OK a maximal ideal (i.e. finite place),and Kp the completion of K at p. The finite adele ring of K (or the finite adelesof K) is the restricted product of all finite places Kp with respect to the family OKp ;that is,

AK,f :=∏p

′Kp ⊆

∏p

Kp,

the prime on the product indicating that, if x = (xp)p ∈ AK,f with every xp ∈ Kp,then xp ∈ OKp for all but finitely many places p. Set theoretically:

AK,f ={

(xp)p ∈∏p

Kp : xp ∈ OKp for all but finitely many p}.

Exercise 29. Show that AK,f is a ring under pointwise addition andmultiplication, and that there is a “diagonal embedding” K ↪→ AK,f .

The restricted product construction endows AK,f with a topology: an open neigh-bourhood of 0 in AK,f is a product

∏p Up, with every Up ⊆ Kp an open neighbourhood

of 0 and all but finitely many Up = OKp . In particular,∏

pOKp is a compact openneighbourhood of 0.

We now define the adele ring (or adeles) of K to be

AK := K∞ × AK,f =∏v

′Kv,

46

the restricted product taken over all (finite and infinite) places v, with the producttopology. The diagonal inclusion obviously extends to an embedding K ↪→ AK .

We can create an analogous construction for function fields: for instance, takeK = C(t) and put OK = C[t]. The maximal ideals of OK take the form t−λ, λ ∈ C,with completion Kλ = C((t − λ)), i.e., the ring of formal Laurent series. There isthen an inclusion

K ↪→∏v∈C

′C((t− v)),

the restricted product taken with respect to the power series rings C[[t− v]].

Exercise 30. Just as K∞ = K ⊗Q R = K ⊗Q Q∞, show that

AK,f = K ⊗Q AQ,f and AK = K ⊗Q AQ.

Theorem 13.1. One has

AQ,f = Q +∏p

Zp;

that is, that if x = (xp)p ∈ AQ,f , then there exist λ ∈ Q (⊆ AQ,f ) andµ ∈

∏p Zp such that x = λ+ µ.

Proof. Let S be the (finite) set of primes such that xp ∈ Zp for all p /∈ S; we willinduct on the order of S. If S = ∅, then x ∈

∏p Zp and we are done.

Otherwise, choose p ∈ S and write

xp =∞∑

m=−n

ampm ∈ Qp,

where a−n 6= 0; put λ = a−np−n + · · · + a−1p

−1, and write λ = Npn

where N is an

integer. Then clearly λ ∈ Z` for all ` 6= p, and so y = (yp)p = x − λ has yp ∈ Zpfor all p /∈ S \ {p}, which has strictly smaller cardinality; so by induction, we aredone.

Exercise 31. Show that AK,f = K +∏

pOKp for a general numberfield K.

We turn our attention for the moment to the group of units of the topologicalring AK , i.e., A×K . This is comparatively small with regard to the ambient ring: for

47

instance, if xp = p for all finite places and x∞ = 1 for the infinite place, then x ∈ AQ,but x−1 /∈ AQ. One can check that

A×K,f =∏p

′K×p ,

the restricted product taken with respect to the subgroups O×Kp, and hence that

A×K =∏v

′K×v .

We call A×K the idele group (or simply the ideles) of K. Again, the restrictedproduct gives us a topology on the ideles, which is not the subspace topology of A×Kin AK .

Remark. We could generalize idele group in the following way: for fixed number n,

GLn(AK) := {(d, g) ∈ AK ×Mn×n(AK)|d det(g) = 1 ∈ AK}

In particular, if n = 1, we have GL1(AK) = A×K .

Both the adeles and the ideles are natural objects in global class field theory.

Theorem 13.2. There exists a continuous, surjective group homomor-phism

K×\A×KrK� Gal(Kab/K),

called the global Artin map, with nontrivial kernel.

We know a priori that the global Artin map cannot be an isomorphism, becauseits domain contains K×∞, while its codomain is profinite. In particular, the image of(K×∞)◦

(i.e. the connected component of the identity) lies in a totally disconnectedgroup, and so must be a singleton.

It turns out that, if CK is the image of(K×∞)◦

in K×\A×K , then ker rK is preciselythe closure of CK . We now list without proof two other properties of rK :

1. For every finite place p of K, there is a commutative diagram

K×\A×K Gal(Kab/K)

K×p Gal(Kp/Kp)ab

rK

rKp

48

where the inclusion Kp ↪→ K×\A×K sends x to (x)p ×∏

v 6=p(1)v.

2. If L/K is a finite extension, then there is a commutative diagram

L×\A×L Gal(Lab/L)

K×\A×K Gal(Kab/K)

rL

N

rK

where N is the norm map.

49

14 Lecture Fourteen

Remark. We know the kernel of the global Artin homomorphism onto Gal(Kab/K),so we “know” the group Gal(Kab/K). In general, however, we don’t know Kab.

In the case where K is Q or imaginary quadratic, we do know Kab (see Serre’sother Cassels-Frohlich article.)

Let’s analyze Q×\A×.

Lemma 2.

A×Q = Q×.

(∏p

Z×p × R>0

),

i.e., if (xv) ∈ A×Q, then there exists λ ∈ Q× such that xp/λ ∈ Z×p forall p, and x∞/λ > 0.

Proof. We have to look at the bad set from (xv) ∈ A×Q, S = {p∣∣xp 6∈ Z×p }. We induct

on the cardinality of S. If S = ∅, then λ = ±1 depending on the sign of x∞. Forgeneral S 6= ∅, then choose some prime p ∈ S, so that xp ∈ Q×p and xp /∈ Z×p .

Rewrite xp = pnu where u ∈ Z×p , and set λ = pn. Then λ ∈ Z×r , for all r 6= p.Dividing by λ, we reduce to the case S\{p}, and by induction we are done.

Remark. In the case when K is a number field, this is hard to push through.Suppose we have some xp, where p is some finite place. What happens if vp(xp) = n?Then we want to set λ ∈ K×, with OKλ = pn.

But what if pn is not principal? Then the lemma is no longer true.

Exercise 32. Show that the double quotient

K×\A×K/

(∏p

O×Kp×K×∞

)is precisely the ideal class group of K.

Remark. In fact, we’ve shown that

A×Q =(Q×)×

(∏p

Z×p × R>0

),

because

Q× ∩

(∏p

Z×p × R>0

)= {1}

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In fact, we’ve gotten something for free!

Corollary 1. Q×\A×Q is canonically isomorphic to∏

p Z×p × R>0.

Corollary 2. ker rQ = R>0

Corollary 3. Gal(Q/Q) is canonically isomorphic to∏p

Z×p = Z× = lim←−n≥1

(Z/nZ)× = Z×,

which of course suggests that Qab is just the maximal cyclotomic ex-tension of Q (which is true.)

Definition. Let K be a number field. A Groessencharacter (also sometimes calleda Hecke character, and we will write it as GC) is a continuous group homomorphismK×\A×K → C×.7

Example 5. Let K = Q. Q×\A×Q = Z× × R>0. If we want to describe the GCs forQ, we just need to figure out what all the continuous group homomorphisms to C×from either component of Z× × R>0.

Exercise 33. Show that the continuous group homomorphisms R>0 →C× are all of the form x 7→ xs

df= exp(s log x).

What about Z×? Note that C× has no “small subgroups”.Let α : Z× → C× be a continuous group homomorphism.Write Z× = lim

←−(Z/nZ)×, so for U open, we have α−1(U) open, hence α−1(U) ⊇

ker(Z× → (Z/nZ)×

)df= KN .

Now, a(Kn) ⊆ U , and is a subgroup. Therefore, α(Kn) = 1 and α factors as

Z× � (Z/NZ)×χ→ C×.

The upshot is: a GC for C is a map Q×\A×Q → C× and for each GC, there exists

a pair (χ, s) where χ is a Dirichlet character and s ∈ C, and a GC on Z× × R>0

factors as Z× × R>0 → (Z/nZ)× × R>0 by (n, x) 7→ χ(n) · xs.7Why do we care about these things? Here’s a secret: GCs are automorphic representations for

GL1(K). (This was Langland’s insight—now we have to figure out how to change the 1 to a 2.)

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Corollary 4. (Due to Tate.) The set of GCs for Q has the structureof a Riemann surface.

Fix a Dirichlet character χ and embed C ↪→ {GCs} by s 7→ (χ, s).In Tate’s thesis, he takes a GC ψ and defines a number L(ψ) ∈

C ∪ {∞}, which induces a complex-valued function L on the Riemannsurface of GCs, which admits a meromorphic extension to all of theC-eigencurve.

Tate checks that the restriction of L to the copy of C attached to χis L(χ, s).

Here’s a generalization of this result to K. Recall for Kp/Qp finite, there’s acanonical norm where the norm for the uniformizer π is 1

q(where q is the size of the

residue field #kp), in terms of scaling the additive Haar measure.The same trick extends to AK ; namely, there exists a canonical Haar measure on

AK for which multiplication by x ∈ A×K scales this Haar measure by ‖x‖, which issome positive real ‖x‖ ∈ R>0.

This therefore defines a norm ‖ · ‖ : A×K → R>0. Since K× diagonally embedsinto A×K , this canonical norm must be trivial on K×.

The upshot is that ‖ · ‖ : K×\A×K → R>0 restricts to the p-adic norm ‖ · ‖p onall Kp, to the usual norm on copies of R×, and to the square of the usual norm oncopies of C×.

So we now have an analogue of what we had for the rational numbers. Hence,the set of all GCs for K also becomes a Riemann surface.

The idea behind the global Langlands philosophy is that there exists a globalLanglands group LQ (in general LK for general K), such that Lab

K = K×\A×K .And now we have shown the 1-dimensional global Langlands correspondence:

there exists a canonical bijection between automorphic representations of GL1(K)with 1-dimensional representations of LK .

Remark. Here’s something to think about: if s ∈ Z, life would be better: K = Qand a GC ψ = (χ, s), where χ is a Dirichlet character, so that E0 = Q(ζm) ⊆ C,so that χ induces χλ : Gal(Q/Q) → GL1((E0)λ), which is a compatible system ofλ-adic Galois representations.

So when s = 1, α is trivial, and ψ = ‖ · ‖−1, ‖πp‖ = 1qp

.

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15 Lecture Fifteen

15.1 Recalling definitions

15.1.1 Compatible systems of λ-adic representations

Recall the following definition. For K,E0 number fields, S a finite set of maximalideals ofOK (a finite set of finite places ofOK), p any finite place of K with p /∈ S andFP(X) a polynomial in E0[X] which is monic of degree n, we define a compatiblesystem of λ-adic Galois representations as a collection of representations

{ρλ : Gal(K/K)→ GLn((E0)λ)},

the index running over all finite places λ of E0, such that: for all ρλ and for all p - `with λ | ` and p /∈ S, the representation ρλ is unramified at p and ρλ(Frobp) hascharacteristic polynomial Fp(X), independent of λ.

We give some examples:

(i) A Dirichlet character χ : (Z/NZ)× → C× in fact has image in E0 = Q(ζd),where d = #(Z/NZ)× = φ(N) (i.e. the Euler totient function). So if we letK = Q, S = {p | N}, we may define ρλ : Gal(Q/Q) → Gal(Q(ζN)/Q) →GL1((E0)λ), which gives a “stupid” example of a compatible system of Galoisrepresentations.

It is immediately apparent that ρλ(Frobp) for p /∈ S does not depend on λ since

it will in fact lie in GL1(E0) ⊆ GL1((E0)λ) and its characteristic polynomial isby definition X − χ(p).

(ii) Consider, for any integer n ∈ Z, the nth power of the cyclotomic characterω` : Gal(Q/Q) → Gal(Q(ζ`∞)/Q) ∼= Z×` ↪→ GL1(Q`). Then for K = Q =E0, S = ∅, Fp(X) = X − p, we see that ω`(Frobp) = p independent of ` forp 6= `.

Then the nth power of the cyclotomic character also gives a compatible systemas desired.

(iii) The Tate module T`E of an elliptic curve E also gives an example.

15.1.2 Grossencharacters

We also will recall the definition of a Grossencharacter. For K = Q we have A×K =

Q× × Z× × R>0; we note that any continuous group homomorphism K×\A×K → C×must factor through Z× × R>0 → (Z/NZ)× × R>0 for some N . For example:

53

(i) Recall that any Grossencharacter for K = Q is defined by a complex numbers and a Dirichlet character (Z/NZ)× → C× such that on the real componentof Q×\A×Q we have x 7→ xs, and on the profinite part the map is given by theDirichlet character.

(ii) Consider the case K = Q(i).

The adeles AK are equal to AK,f ×K∞. In this case, we have an isomorphismK∞ ∼= C. We may define A×K,f =

∏′pK

×p , the restricted product indicating that

if (xp)p ∈ A×K,f , then vp(xp) = 0 for all but finitely many p.

Note the following construction: given a finite idele x = (xp)p, define np :=vp(xp) ∈ Z for each p. Clearly np = 0 for all but finitely many p.

Define a fractional ideal I(x) =∏

p pnp ⊆ K. This is in fact a finite product,

which is an ideal if all np ≥ 0.

Thus we get a map from A×K,f to the set of fractional ideals of K, under whichthe image of K× is the set of principal ideals of K – so, the quotient maps tothe class group, in some sense.

We return to our case K = Q(i). We want to write down a Grossencharacter.

We see that A×K = A×K,f ×K×∞. We claim that A×K = K× · (∏

pO×Kp×K×∞). This

is a consequence of the fact that K has class number 1.

Proof. Write x ∈ A×K as a product xf × x∞ where xf is a finite idele andx∞ ∈ C×.

Then since K has class number 1, I(xf ) = (λ) for some λ ∈ K×. Then one cancheck easily that vp(xp/λ) = 0 for all p; hence xf/λ ∈

∏pO×Kp

, as desired.

This proof works for any K with class number 1.

Then for any K a number field of class number 1 we see that to give aGrossencharacter ψ : K×\A×K → C× it suffices to define a continuous ψ onK×∞ ×

∏pO×Kp

, such that ψ is trivial on K× ∩ (∏

pO×Kp×K×∞) = O×K .

Thus for K = Q(i), take any ideal N ⊆ Z[i]. We then take any group homo-morphism χ : (Z[i]/N )× → C×.

Then χ gives us a map∏

pO×Kp

= (OK)× → (OK/N )× → C×.

Abstractly as a group, R>0 × S1 ∼= C×. We can define a group homomorphismon the real part by x 7→ xs for some s ∈ C and on the S1-part by z 7→ zn forsome n ∈ Z. This exhausts the group homomorphisms C× → C×.

54

The upshot so far is that given such χ, s, n, we get some ψ0 :∏

pO×Kp×K×∞ →

C×.

However, ψ0 may not extend to a Grossencharacter. When will it? Exactlywhen χ and C× → C× given by s, n are both trivial on O×K = {±1,±i}.

Exercise 34. With ψ0 defined as above, check that ψ04 always

extends to a Grossencharacter.

(iii) Now we will consider K = Q(√

2). We note that OK = Z[√

2] and O×K =±1× 〈1 +

√2〉.

Here A×K = K× · (∏

pO×Kp×K×∞) since we again have class number 1.

A character of∏

pO×Kp

will factor through some map α : (Z[√

2]/N )|times →C× (same argument with profinite completion).

In this case K×∞∼= R× × R×. A character will look like

χ∞(x1, x2) = |x1|s1|x2|s2 sgn(x1)e1 sgn(x2)e2 ,

for two complex numbers si and a choice of sign ei ∈ {0, 1} for each i = 1, 2.

We can then define ψ0 :∏

pO×Kp×K×∞ → C×.

How do we get ψ0 to extend to a Grossencharacter? This happens when ψ0(1+√2) is trivial - then |1 +

√2|s1|1−

√2|s2 is a root of unity. If 1 +

√2 = x, then

|1 −√

2| = 1/x; hence it is clear that we must have xs1−s2 a root of unity, sos1 − s2 must be in πi

log(1+√

2)Q, which is a discrete set.

The upshot is that if we have χ : (Z[√

2]/N )× → C× and two complex numberss1, s1 + q for q in some discrete set, and some signs, then the corresponding ψ0

to some finite power does extend to a Grossencharacter.

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16 Lecture Sixteen

16.1 Global Langlands Conjecture for GL1

We’ve seen examples of compatible systems of one-dimensional Galois representationsand GCs.

Definition. A GC ψ : K×\AK× → C× is said to be algebraic if its restriction to the

connected component of the identity (K∞×)o ' (R>0)r × Cs is defined by

ψ(x1, . . . , xr, z1, . . . , zs) = xn11 · · ·xnrr z

nr+1

1 znr+2

1 · · · znr+2s−1s znr+2s

s ,

with each ni ∈ Z.

Example 6. 1. ‖ · ‖ : K×\AK× → C× is algebraic with all ni = 1.

2. For K = Q the GC x 7→ xs is algebraic if and only if s ∈ Z.

Philosophy: Let χ be a GC for K (an automorphic representation for GL1/K); itcorresponds to one-dimensional representations of the (conjectural) global Langlandsgroup LK . Since we can’t even define the global Langlands group, this correspon-dence is not so much a conjecture as it is a philosophy.

Theorem 16.1. (Weil, Basic Number Theory?) Let χ be an algebraicGC, then there exists a compatible system of λ-adic Galois representa-tions attached to χ, and conversely.

Idea: Let ψ : K×\AK× → C×, then ψ|(K∞×)o : x 7→ xn. We want a representation

Gal(K/K)ab = K×\AK×/(K∞

×)◦ → GL1(Q`).

Proof. Given an algebraic GC, χ, define

χ|(K∞×)o(x∞) =∏v real

xnvv ×∏

v↔{σ,c◦σ}

(σxv)nv,1(σxv)

nv,2

with nv ∈ Z, and the second product taken over the complex embeddings σ, c ◦ σ :K → C (as usual, c is complex conjugation).

We further define

χ0 : AK× → C×

x 7→ χ(x)

/( ∏v real

xnvv ×∏

v↔{σ,c◦σ}

(σxv)nv,1(σxv)

nv,2

).

56

We note that χ0 is trivial on the connected component but is non-trivial on K×

(but earlier it was trivial). For λ ∈ K×, χ0(λ) =∏

σ:K→C σ(λ)nσ where nσ ∈ Z, thusnon-trivial.

One checks that Im(χ0) ⊆ E0 (a number field). Say λ is a finite place of E0, so thatthe image of

χ0|K× : Q× → Q`×

µ 7→∏σ

σ(µ)nσ

lies in E0 ⊂ E0,λ with nσ ∈ Z. It extends to a continuous group homomorphism

χ` : (K ⊗Q Q`)× =

∏p|`

Kp× → (E0,λ)

×.

We further define, for λ | `,

ψλ : AK× → (E0,λ)

×

x 7→ χ0(x)

χ`(x`)

which is very complicated at Kp for p | `, but is ψλ = χ at all places p - `.Since ψλ is trivial on the connected component and onK×, it extends to Gal(K/K)ab =

K×\AK×/(K∞

×)◦.

Fp(X) = X − χ0(πKp) for a uniformizer πKp and hence it is a compatible system.

Conversely, if ψλ is a compatible system, to show that it comes from an algebraicGC we use Waldschmidt’s theorem from transcendence theory.

Remark. The big picture:{automorphic representa-tions of GL1/K

}←→

{one-dimensional rep-resentations of LK

}{

algebraic automorphic rep-resentations of GL1/K

}←→

{compatible systems ofone-dimensional repre-sentations of Gal(K/K)

}{p-adic automorphic rep-resentations of GL1/K

}←→

{continuous p-adic repre-sentations Gal(K/K) →GL1(Qp)

}

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The second correspondence is the theorem we just proved above. There is an n-dimensional analogue of the above statements, where 1 is replaced by n in each case.However, for general n ≥ 1, most of the terms in this picture aren’t even well-definedand, hence it’s just a philosophy.

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17 Lecture Seventeen

Last time, we discussed a series of correspondences related to the global Lang-lands program; one was the correspondence between the set of algebraic automor-phic representations for G over K (where G is any connected, reductive algebraicgroup), and the set of compatible systems of semisimple `-adic Galois representationsGal(K/K)→ LG(Qp) (i.e. the Langlands dual group).

In the case G = GLn, we will have LG = GLn, although for general G there aresome subtleties. One is that there is more than one notion of algebraicity, two ofwhich we might call C-algebraic and L-algebraic, respectively; we will discuss thesemore below. Another issue is that the correspondence may fail to be a bijection forgeneralG, and may instead be a finite-to-one correspondence; this arises from the factthat the L-packets, introduced above, are collections of automorphic representationswhich may still give the same `-adic representation ρ. A further issue arises fromthe fact that different global Langlands parameters may be isomorphic everywherelocally.

These obstacles are not insurmountable; however, they complicate the under-standing of our “correspondence,” which must then be investigated more carefully.

Philosophically, we want the `-adic representation ρπ associated to a given au-tomorphic representation π of G to be defined only up to some Tate-Shafarevichgroup, which is trivial when G = GLn, but not generally. We will not pursue theseconsiderations deeply, and will restrict our attention for the moment to the groupG = GLn.

The correspondence in this case was worked out in a series of papers by Langlands-Corvallis, and further by Clozel. Explicit correspondences were worked out in thecase when π is an automorphic representation of GLn over K, where K is totallyreal (or has complex multiplication), and satisfies both a certain strong self-dualitycondition, and a strong algebraicity (or cohomological) condition; Clozel was able towrite down precisely a compatible system ρπ.

Clozel’s proof goes roughly as follows: given π, we find an appropriate Shimuravariety X, then relate the cohomology of X to automorphic forms, using the Eichler-Shimura congruence relation. More recently (c. 2013), a paper by Harris-Lan-Taylor-Thorne were able to remove the self-duality condition by (roughly speaking) consid-ering the representation π ⊕ π∨ rather than π itself. Taking limits of cohomology ofShimura varieties yields the desired compatible system ρ.

These ideas are rooted in Weil’s construction (which we saw yesterday) of an ap-propriate one-dimensional `-adic representation ρχ attached to a given Großencharak-ter χ. Eichler, Shimura, Deligne, and Serre were able to show that, if f is a weight k

59

modular eigenform, then there exists a compatible system of two-dimensional Galoisrepresentations Gal(Q/Q).

Now, before we introduce the notion of an automorphic representation, we willinvestigate what properties these representations will not satisfy. Recall that thelocal Langlands conjectures concern themselves with the set of all smooth-admissible,irreducible representations of GLn(K) for a finite extension K/Qp, and the set ofcertain n-dimensional Weil-Deligne representations.

By contrast, the global Langlands conjectures will be concerned with automorphicrepresentations of GLn(AK), where K is a number field; by definition, an automor-phic representation will be irreducible. It is not hard to show that

GLn(AK) =∏v

′GLn(Kv),

the restricted product taken with respect to the subgroups GLn(OKv). By Flath’s the-orem, we deduce that certain irreducible, well-behaved representations of GLn(AK)correspond to tensor products of representations of the factors GLn(Kv); we willwrite

π =⊗v

′πv,

each πv an irreducible, admissible representation of GLn(Kv).The idea is now as follows: if π is such a representation of GLn(AK), correspond-

ing to the compatible system ρ under the global Langlands correspondence, thenfor every finite place, the representation πv will correspond (by the local Langlandscorrespondence) to a Weil-Deligne representation ρ0,v : Gal(Kv/Kv) → GLn(Q`),obtained by restriction.

Consequently: the definition of an automorphic representation of GLn overK can-not simply be an arbitrary smooth-admissible irreducible representation of GLn(AK);here the analogy with the local Langlands correspondence begins to fall apart.

To illustrate this point, we consider an example in dimension one. Suppose on thecontrary that an automorphic representation of GL1/K is simply a representation ofA×K , and let K = Q. By Flath’s theorem it suffices to give representations of Q×v forevery place v.

For the first twenty-five primes p (say), let us define our representation of Q×p viaZ×p 7→ 1 and p 7→ 7; for the remaining primes we will take the trivial representation,and our representation of Q×∞ = R× will also be trivial. Finally, we put

π =⊗v

′πv,

60

as above, and we have a completely valid representation π of GL1(AQ). If π is in-deed automorphic, then it corresponds (under global Langlands) to some compatiblesystem

ρ` : Gal(Q/Q)→ GL1(Q`),

which we want to respect the local Langlands correspondence. In particular, this willmean that ρ`(Frobp) = 1 for all p > 97, and so by the Chebotarev density theoremwe must have that ρ` is the trivial one-dimensional representation. But this impliesthat

ρ`(Frob2) = 1

and not seven, as we have demanded. Thus we see that we will need to take a morerestrictive definition of an automorphic representation, in stark contrast to the localcase.

The main failure of our pathological example π is that it will fail to be trivial onQ×, as is the case for Großencharakteren; indeed, we have

π(2) = π(2∞, 22, 23, 25, . . .) = 1 · 7 · 1 · 1 · · · = 7 6= 1.

Now: if G is a finite group, suppose we want to find all irreducible representationsof G. Clearly it will suffice to investigate the group ring

C[G] ∼=⊕

π irreducible

πdimπ.

In fact, if H ≤ G is a subgroup, it is enough to look at C[H\G], which we identifywith the set of functions H\G → C on the coset space. We may view C[H\G] asa subspace of C[G], and there will be some set S (generally not the entire set ofirreducible representations) and integers m(π) such that

C[H\G] ∼=⊕π∈S

πm(π) ⊆ C[G].

Generalizing this idea: taking instead G = GLn(AK), we consider the functionsGLn(K)\GLn(AK) → C, satisfying certain (not too restrictive) conditions – callthese the nice functions φ, and let A0 = A0

(GLn(K)\GLn(AK)

)denote the vector

space of all such nice φ. There is an action of GLn(AK) on A0 via

(g.φ)(γ) = φ(γg).

Exercise 35. Check that this is indeed a group action, and is well-defined in the sense that (g.φ)(αγ) = (g.φ)(γ) for α ∈ GLn(K).

61

In the case n = 1, a Großencharakter will be a nice function in A0, and indeed,a finite sum of arbitrary Großencharakteren will also be nice.

In general, the space A0 will not be irreducible, but may turn out to be a directsum of irreducible representations π of GLn(AK); thus we are motivated to investigatethese representations π, and we consider these possible candidates for the definitionof an automorphic representation. We will make this more precise in the next lecture.

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18 Lecture Eighteen

Let K be a number field and let S be a finite set of finite places.Consider Gal(KS/K). We have conjugacy classes Frobp for each p 6∈ S. If

Gal(KS/K) happened to be a free group, freely generated by these Frobp’s, then wecould just send Frobenius anywhere we like, i.e. we can define ρ : Gal(KS/K) →GLn(Q`) by choosing random matrices ρ(Frobp) = Mp ∈ GLn(Q`) for all p 6∈ S, andthen we’re finished.

And then the global Langlands conjectures would say: “take any old irreduciblerepresentation π of GLn(AK) and we’ll get ρπ.”

How this would work is that we’d have a decomposition π =⊗

πv, sending πpunramified to Mp.

But this is not the truth, because Gal(KS/K) is not freely generated by theFrobp, and these Frobp are all related in some vastly complex way which no humanunderstands.

For example, we could just let K be the rational numbers and let S be the primes{2, 3, 5}. And in this case we could ask: could we choose a member of the Frob7-conjugacy class and a member of the Frob11-conjugacy class which multiply to theFrob23-conjugacy class? Things like that.

But! The Chebotarev density theorem gives us something. It says that the Frobp

are dense. In fact it tells us that if we remove a single Frobp, then what’s left is stilldense, so it tells us that this single one that we removed can be written as a limit ofthe others.

So it’s complicated, and figuring out these relations is really hard.So on the automorphic representation side, we don’t get this nice tensor product

decomposition, because this would correspond to having this freely-generated storyon the Gal(KS/K)-side.

Here’s another approach. Based on our successes for GL1, we will restrict torepresentations π of GLn(AK) which show up in A0 (GLn(K)) which will be “nicefunctions” GLn(K)\GLn(AK)→ C.

Question 1. What’s a nice function?

Well, when n = 1, Groessencharacters were nice. Just compose by the inclusionC× ↪→ C.

Let χ be a Groessencharacter, χ : GL1(AK)→ C.Define (g ∗ χ)(γ) = χ(γg) = χ(g)χ(γ), and therefore

g ∗ χ = χ(g) · χ.

63

Hence Cχ is a one-dimensional vector space on which GL1(AK) acts via χ.Also, Groessencharacters are locally constant at finite places and smooth at x.

They’re of the form x 7→ xs. In fact, they’re more than smooth. They’re alsogroup homomorphisms. But that doesn’t help us here, because nice functions outof GLn(K)\GLn(AK) are functions on a coset space, which a priori might not be agroup (because GLn(K) is generally not a normal subgroup).

Let’s differentiate f(x) = xs = exp(s log x), so f ′(x) = exp(s log x) · sx

= xs · sx.

Therefore, xf ′(x) = sxs = sf(x).So it satisfies this natural differential equation.We need machinery which spits out differential equations.

Definition. “Nice” means:

• locally constant at finite places

• left GLn(K)-invariant

• smooth at infinite places

• satisfies a certain differential equation

• boundedness

18.1 Interlude on differential equations

A Lie group is a group object in the category of smooth manifolds. Given a Liegroup G, we can form its Lie algebra g, which is the tangent space of the identity; gcan be thought of as comprising “differential operators on G”.

There is a natural exponential map g→ G.If we have an X ∈ g, then we view this as inducing a differential operator on the

space of{C∞-functions G→ C}

via

(X ∗ f)(g)df=

d

dt(f(g · exp(tX)))

∣∣∣∣t=0

.

Let’s see an example of this. Let G = GL1(R) = R×.Then g = R, so let X = 1. Then

(X ∗ f) (g) =d

dt

(f(g · et)

) ∣∣∣∣t=0

,

64

and differentiating, we obtainf ′(g)g.

The point here is that there is a canonical differential operator coming fromthe Lie algebra, and this canonical differential operator is apparently not the usualderivative. To recap, we have a differential equation

(X ∗ f)(x) = xf ′(x), x ∈ R×.

So, returning to our differential equation from before

xf ′(x) = sxs,

we then want to write X ∗ f = s · f .What does this look like for GL2(R)? In this case g = M2(R) has basis

E =

(0 10 0

), F =

(0 01 0

), H =

(1 00 1

), Z =

(0 11 0

).

Each of these gives a differential operator on the vector space of C∞-functionsGL2(R)→ C; call the space of such functions V .

Now, we have operators E,F,H,Z : V → V , and these operators do not commute(e.g. EF − FE = H. So it’s a bad idea to ask for simultaneous eigenfunctions.

So here’s the plan: we’re going to find a bunch of differential operators thatcommute!

Let’s start with a Lie algebra g (keep in mind gl2(R)), which is some finite-dimensional vector space g over R, equipped with a Lie bracket [·, ·] : g× g→ g.

Definition. The universal enveloping algebra of a Lie algebra g is the associativealgebra U(g) such that the functor g 7→ U(g) is left-adjoint to the forgetful functorfrom the category of unital associative algebras to the category of Lie algebras (i.e. thefunctor which takes the underlying vector space and attaches the commutator asbracket, [X, Y ] := XY − Y X).

Let g be a Lie algebra with R-basis {X1, . . . , Xd}. This contains all the first orderdifferential operators. The associative algebra R〈X1, . . . , Xd〉 of non-commutativepolynomials in d variables contains (non-commuting) higher order differential oper-ators. Taking the quotient by the bi-ideal generated by relations (XiXj −XjXi) −[Xi, Xj], we obtain the universal enveloping algebra U(g).

What’s the centre of this algebra? It’s a whole bunch of commuting differen-tial operators, for which we might hope that our nice functions are simultaneouseigenforms.

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Harish-Chandra figured out what Z(U(gC)) is, where gC = g⊗R C.g contains a Cartan subalgebra h, and U(hC) is the commutative polynomial ring

C[X1, . . . , Xd].

Theorem 18.1. (Harish-Chandra homomorphism) Let g = gln(R).Then, there exists a canonical injection

Z (U(gC)) ↪→ U(hC),

and the image is U(gC)W , where W is the Weyl group.

Specializing to n = 2, we get get gC = CE ⊕ CF ⊕ CZ ⊕ CH, with U(gC) 3EF + FFH, and

Z(U(gC)) ' U(C⊕ C)W = C[X, Y ]Z/2Z.

So now we have a problem: what are the polynomials in 2 variables which arethe same when we interchange X and Y ?

Exercise 36. Show that

Z(U(gC)) ' U(C⊕ C)W = C[X, Y ]Z/2Z = C[S, T ],

where S = XY and T = X + Y .

Tomorrow8 we’ll see the definition of an automorphic representation.

19 Lecture 19

For G a connected reductive group over K (in most cases G = GLn), we want todefine A(G/K) = {φ : G(K)\G(AK)|φ nice} for some value of “nice”.

These “nice” functions will be automorphic forms.As in the GL1 case, where GL1(AK) ⊇ K×

∏P O

×KPK×∞, which is a subgroup of

finite index (quotient by which is the class group).φ has to be trivial on K×, and on

∏P O

×KP

is in practice finite, with K×∞ a realmanifold.

The same story is true in the GLn case. The proof for GL2/Q is coming later.

GLn(AK) ⊇ GLn(K)∏P

GLn(OKP )GLn(K∞)

8“When are we leaving again?” (Note: tomorrow’s the last day.)

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as a subgroup with finite index; again, φ has to be trivial on GLn(K), and we geta finite amount of information on GLn(OKP ), with GLn(K∞) some real manifoldagain.

As a reminder, in the case GL1/Q, at ∞ any φ is a function on R×/Z×, and wewant Grossencharacters to exist in A(GL1/Q).

The conclusion we get from this is that R>0 → C× given by x 7→ xs must be nice.There is some canonical connection with differential operators in this situation.

For G(R) the Lie group, we get a Lie algebra g; for the GL1-case, we can choosea bvasis vector D of g so that (Df)(x) = xf ′(x), and D(x 7→ xs) = x 7→ sxs, sox 7→ xs is an eigenfunction of D - annihilated by D − s.

However, we also want functions like f(x) = xs + 7xt to be ‘nice’ so that A isa vector space. This function is not annihilated by D − s, but it is annihilated by(D − s)(D − t).

Abstractly, the algebra C[D] acts on C∞ functions R>0 → C, with sums ofGrossencharacters annihilated by a nonzero idea in C[D].

If I = {D′ ∈ C[D] : D′(f) = 0} for f a sum of Grossencharacters, this is evidentlyan ideal. In the cases we have seen so far, I had finite codimension; we see that if wedefine a map C[D]→ C2 by h(D) 7→ (h(s), h(t)), the kernel is I = ((D− s)(D− t)),so I has codimension 2.

For general G, we have Lie algebra g of G(K∞) with basis E1, E2, . . . , Ed of g -however, these basis elements do not commute.

Instead we consider the universal enveloping algebra U(gC) as defined before(C〈E1, . . . , Ed〉/(EiEj−EjEi− [Ei, Ej])i,j, the quotient of the noncommutative poly-nomial algebra in the basis elements by the bi-ideal given by the relations EiEj −EjEi = [Ei, Ej] for the Lie bracket [, ] of g). Harish-Chandra showed that forG = GLn/Q the center Z(U(gC)) = C[T1, T2, . . . , Tn]W where W is the symmet-ric group Sn acting on the Ti in the usual way.

The upshot is that Z(U(gC)) is a canonical source of higher-order differentialoperators. Alternatively we can think of these as bi-G-invariant differential operators.

We have a torus T in GLn/Q given by

(∗ 0 ... 00 ∗ ... 0... ... ... ...0 0 ... ∗

)⊆ GLn. Then we get a Lie

subgroup T (R) ⊆ G(R), with Lie subalgebra t ⊆ g.t has basis t1, . . . , tn and U(t) = C[t1, . . . , tn], with W = NG(T )/T the Weyl group

in this case isomorphic to Sn with the usual action on the ti.Then Z(U(g)) = C[t1, . . . , tn]Sn = C[σ1, . . . , σn] for σi the ith symmetric polyno-

mial.For example, in the case GL2/Q, this machine tells us that Z(U(g)) = C[∆, Z]

for two differential operators ∆, Z.

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In this case note that σ1 = t1 + t2, σ2 = t1t2.If g = Lie(GL2(R)), it is Mat2(R) with the commutator Lie bracket. Recall its

basis {E = ( 0 10 0 ) ,F = ( 0 0

1 0 ) ,H = ( 1 00 −1 ) ,Z = ( 1 0

0 1 )}. The bracket is given by[E ,F ] = H, [E ,H] = −2E ,F ,H = 2F , [∗,Z] = 0.

Then U(gC) = C〈E,F,H,Z〉/(EF − FE − H,EH − HE + 2E,FH − HF −2F, etc.).

Exercise: If ∆ := H2 + 2EF + 2FE, ∆ commutes with E,F,H,Z.Then ∆, Z ∈ Z(U(gC)). It will turn out that they in fact generate the center -

Z(U(gC) = C[∆, Z].∆ is some second order differential operator on {f : GL+

2 (R) → C}. As areminder, GL+

2 (R) (matrices with positive determinant) acts on the upper half planeh by ( a bc d ) τ = aτ+b

cτ+d; this action is transitive. Then there is a surjection GL+

2 (R)→ hvia γ 7→ γi.

The stabilizer of i isR×SO2(R). Then the upper half plane is GL+2 (R)/(R×SO2(R))

So now for f : h → C and F the associated function on GL+2 (R), the desired

invariance properties for F carry over to ∆F and we get descent to a function ∆fon h.

What is this operator? Up to a constant, ∆f = −y2(∂2f∂x2

+ ∂2f∂y2

). The Cauchy-Riemann equations imply that holomorphic and antiholomorphic functions are an-nihilated by this.

For GL1/Q our definition of nice was that there was some nonzero ideal in C[D]annihilating f .

Then once again we want f such that some finite codimension ideal in C[∆, Z] =Z(U(gC)) annihilates f .

It appears we are interested in functions f : h→ C such that ∆, Z act by scalarsλ, µ respectively on f .

Recall a Theorem of Deligne et al. saying that f a modular eigenform correspondsto ρf : Gal(Q/Q)→ GLn(Q`) with det ρf = −1.

Now let f(x) be some random irreducible cubic polynomial over Q with threereal roots α, β, γ. Let K = Q[α, β, γ]. Chances are Gal(K/Q) ∼= S3.

There is some standard irreducible two-dimensional representation S3 → GL2(Q).Fix ρ0 : Gal(Q/Q) factoring through this map and the surjection to Gal(K/Q). Thenwe find that the determinant of ρ0(x 7→ x) is −1.

For any ` prime, this gives ρ` : Gal(Q/Q) → GL2(Q`). This gives a “bogus”compatible system of `-adic representations.

Attached to this system should be an automorphic representation π by the Lang-lands philosophy; Maas solved this problem by writing down a specific function onthe upper half plane to the complex numbers that is not holomorphic but is invariant

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under some Γ1(N) for N the conductor of ρ0, and ∆f = λf for some λ 6= 0.

Definition. Take G a connected reductive group over a number field K, a fixedmaximal compact subgroup H∞ (sometimes denoted K∞) of G(K∞). (For example,in our usual examples G = GLn/Q, G(K∞) = GLn(R), H∞ = On(R).) We call afunction φ : G(K)\G(AK)→ C is called an automorphic form if

1. φ is smooth - i.e., if we write G(AK) = G(AK,f )×G(K∞) and write elementsas (x, y), then for x fixed, φ is C∞ with regard to y, and for fixed y, φ is locallyconstant with regard to x.

2. φ is well-behaved on compact subgroups in a way somehow related to admissi-bility - i.e.

(a) There exists Uf ⊆ G(AK,f ) compact open such that φ(gu) = φ(g) for anyu ∈ Uf ;

(b) the C-vector spaced spanned by functions of the form g 7→ φ(gh∞) for eachh∞ ∈ H∞ is finite-dimensional (e.g. φ trivial on H∞).

3. There exists some ideal I ⊆ Z(U(gC)) for g the Lie algebra of G(K∞) suchthat I has finite codimension and for any differential operator D ∈ I we haveD(y 7→ φ(x, y)) = 0 for all x ∈ G(AK,f ).

4. Some “boring” growth condition: |φ(x, y)| is at most some constant times ||y||Nfor some N and some sensible norm on G(K∞).

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